Parabolas

Graph Vertical Parabolas

The next conic section we will look at is a parabola. We define a parabola as all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

Previously, we learned to graph vertical parabolas from the general form or the standard form using properties. Those methods will also work here. We will summarize the properties here.

Vertical Parabolas
	General form \(y=a{x}^{2}+bx+c\)	Standard form \(y=a{(x-h)}^{2}+k\)
Orientation	\(a>0\) up; \(a<0\) down	\(a>0\) up; \(a<0\) down
Axis of symmetry	\(x=-\frac{b}{2a}\)	\(x=h\)
Vertex	Substitute \(x=-\frac{b}{2a}\) and solve for y.	\((h,k)\)
y-intercept	Let \(x=0\)	Let \(x=0\)
x-intercepts	Let \(y=0\)	Let \(y=0\)

The graphs show what the parabolas look like when they open up or down. Their position in relation to the x- or y-axis is merely an example.

To graph a parabola from these forms, we used the following steps.

The next example reviews the method of graphing a parabola from the general form of its equation.

Example

Try it.

Graph \(y=\text{-}{x}^{2}+6x-8\) by using properties.

Solution

Since a is \(-1,\) the parabola opens downward.
To find the axis of symmetry, find \(x=-\frac{b}{2a}.\)
	The axis of symmetry is \(x=3.\)
The vertex is on the line \(x=3.\)
Let \(x=3.\)
	The vertex is \((3,1).\)
The y-intercept occurs when \(x=0.\)
Substitute \(x=0.\)
Simplify.
	The y-intercept is \((0,-8).\)
The point \((0,-8)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,-8).\)	Point symmetric to the y-intercept is \((6,-8).\)
The x-intercept occurs when \(y=0.\)
Let \(y=0.\)
Factor the GCF.
Factor the trinomial.
Solve for x.
	The x-intercepts are \((4,0),(2,0).\)
Graph the parabola.

The next example reviews the method of graphing a parabola from the standard form of its equation, \(y=a{(x-h)}^{2}+k.\)

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Graph Horizontal Parabolas

Our work so far has only dealt with parabolas that open up or down. We are now going to look at horizontal parabolas. These parabolas open either to the left or to the right. If we interchange the x and y in our previous equations for parabolas, we get the equations for the parabolas that open to the left or to the right.

Horizontal Parabolas
	General form \(x=a{y}^{2}+by+c\)	Standard form \(x=a{(y-k)}^{2}+h\)
Orientation	\(a>0\) right; \(a<0\) left	\(a>0\) right; \(a<0\) left
Axis of symmetry	\(y=-\frac{b}{2a}\)	\(y=k\)
Vertex	Substitute \(y=-\frac{b}{2a}\) and solve for x.	\((h,k)\)
y-intercepts	Let \(x=0\)	Let \(x=0\)
x-intercept	Let \(y=0\)	Let \(y=0\)

The graphs show what the parabolas look like when they to the left or to the right. Their position in relation to the x- or y-axis is merely an example.

Looking at these parabolas, do their graphs represent a function? Since both graphs would fail the vertical line test, they do not represent a function.

To graph a parabola that opens to the left or to the right is basically the same as what we did for parabolas that open up or down, with the reversal of the x and y variables.

Example

Try it.

Graph \(x=2{y}^{2}\) by using properties.

Solution

Since \(a=2,\) the parabola opens to the right.
To find the axis of symmetry, find \(y=-\frac{b}{2a}.\)
	The axis of symmetry is \(y=0.\)
The vertex is on the line\(y=0.\)
Let \(y=0.\)
	The vertex is \((0,0).\)

Since the vertex is \((0,0),\) both the x- and y-intercepts are the point \((0,0).\) To graph the parabola we need more points. In this case it is easiest to choose values of y.

We also plot the points symmetric to \((2,1)\) and \((8,2)\) across the y-axis, the points \((2,-1),\)\((8,-2).\)

Graph the parabola.

In the next example, the vertex is not the origin.

In , we see the relationship between the equation in standard form and the properties of the parabola. The How To box lists the steps for graphing a parabola in the standard form \(x=a{(y-k)}^{2}+h.\) We will use this procedure in the next example.

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Solve Applications with Parabolas

Many architectural designs incorporate parabolas. It is not uncommon for bridges to be constructed using parabolas as we will see in the next example.

Example

Try it.

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

Solution

We will first set up a coordinate system and draw the parabola. The graph will give us the information we need to write the equation of the graph in the standard form\(y=a{(x-h)}^{2}+k.\)

Let the lower left side of the bridge be the origin of the coordinate grid at the point \((0,0).\) Since the base is 20 feet wide the point \((20,0)\) represents the lower right side. The bridge is 10 feet high at the highest point. The highest point is the vertex of the parabola so the y-coordinate of the vertex will be 10. Since the bridge is symmetric, the vertex must fall halfway between the left most point, \((0,0),\) and the rightmost point \((20,0).\) From this we know that the x-coordinate of the vertex will also be 10.
Identify the vertex, \((h,k).\)	\((h,k)=(10,10)\)
	\(h=10,\ k=10\)
Substitute the values into the standard form. The value of a is still unknown. To find the value of a use one of the other points on the parabola.	\(\ \begin{array}{lll}y & = & a{(x-h)}^{2}+k \\ y & = & a{(x-10)}^{2}+10 \\ (x,y) & = & (0,0)\end{array}\)
Substitute the values of the other point into the equation.	\(\ \begin{array}{lll}y & = & a{(x-10)}^{2}+10 \\ 0 & = & a{(0-10)}^{2}+10\end{array}\)
Solve for a.	\(\ \begin{array}{lll}0 & = & a{(0-10)}^{2}+10 \\ -10 & = & a{(-10)}^{2} \\ -10 & = & 100a \\ \frac{-10}{100} & = & a \\ a & = & -\frac{1}{10}\end{array}\)
	\(y=a{(x-10)}^{2}+10\)
Substitute the value for a into the equation.	\(\ y=-\frac{1}{10}{(x-10)}^{2}+10\)

Key Concepts

• Parabola: A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.
  

  Vertical Parabolas
  	General form \(y=a{x}^{2}+bx+c\)	Standard form \(y=a{(x-h)}^{2}+k\)
  Orientation	\(a>0\) up; \(a<0\) down	\(a>0\) up; \(a<0\) down
  Axis of symmetry	\(x=-\frac{b}{2a}\)	\(x=h\)
  Vertex	Substitute \(x=-\frac{b}{2a}\) and solve for y.	\((h,k)\)
  y- intercept	Let \(x=0\)	Let \(x=0\)
  x-intercepts	Let \(y=0\)	Let \(y=0\)

• How to graph vertical parabolas \((y=a{x}^{2}+bx+c\) or \(f(x)=a{(x-h)}^{2}+k)\) using properties.
  1. Determine whether the parabola opens upward or downward.
  2. Find the axis of symmetry.
  3. Find the vertex.
  4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
  5. Find the x-intercepts.
  6. Graph the parabola.
  
  

  Horizontal Parabolas
  	General form \(x=a{y}^{2}+by+c\)	Standard form \(x=a{(y-k)}^{2}+h\)
  Orientation	\(a>0\) right; \(a<0\) left	\(a>0\) right; \(a<0\) left
  Axis of symmetry	\(y=-\frac{b}{2a}\)	\(y=k\)
  Vertex	Substitute \(y=-\frac{b}{2a}\) and solve for x.	\((h,k)\)
  y-intercepts	Let \(x=0\)	Let \(x=0\)
  x-intercept	Let \(y=0\)	Let \(y=0\)

• How to graph horizontal parabolas \((x=a{y}^{2}+by+c\) or \(x=a{(y-k)}^{2}+h)\) using properties.
  1. Determine whether the parabola opens to the left or to the right.
  2. Find the axis of symmetry.
  3. Find the vertex.
  4. Find the x-intercept. Find the point symmetric to the x-intercept across the axis of symmetry.
  5. Find the y-intercepts.
  6. Graph the parabola.

Parabolas

Graph Vertical Parabolas

In the following exercises, graph each equation by using properties.

Try it.

\(y=\text{-}{x}^{2}+4x-3\)

Solution

Try it.

\(y=\text{-}{x}^{2}+8x-15\)

Try it.

\(y=6{x}^{2}+2x-1\)

Solution

Try it.

\(y=8{x}^{2}-10x+3\)

In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.

Try it.

\(y=\text{-}{x}^{2}+2x-4\)

Solution

ⓐ \(y=\text{-}{(x-1)}^{2}-3\)
ⓑ

Try it.

\(y=2{x}^{2}+4x+6\)

Try it.

\(y=-2{x}^{2}-4x-5\)

Solution

ⓐ \(y=-2{(x+1)}^{2}-3\)
ⓑ

Try it.

\(y=3{x}^{2}-12x+7\)

Graph Horizontal Parabolas

In the following exercises, graph each equation by using properties.

Try it.

\(x=-2{y}^{2}\)

Solution

Try it.

\(x=3{y}^{2}\)

Try it.

\(x=4{y}^{2}\)

Solution

Try it.

\(x=-4{y}^{2}\)

Try it.

\(x=\text{-}{y}^{2}-2y+3\)

Solution

Try it.

\(x=\text{-}{y}^{2}-4y+5\)

Try it.

\(x={y}^{2}+6y+8\)

Solution

Try it.

\(x={y}^{2}-4y-12\)

Try it.

\(x={(y-2)}^{2}+3\)

Solution

Try it.

\(x={(y-1)}^{2}+4\)

Try it.

\(x=\text{-}{(y-1)}^{2}+2\)

Solution

Try it.

\(x=\text{-}{(y-4)}^{2}+3\)

Try it.

\(x={(y+2)}^{2}+1\)

Solution

Try it.

\(x={(y+1)}^{2}+2\)

Try it.

\(x=\text{-}{(y+3)}^{2}+2\)

Solution

Try it.

\(x=\text{-}{(y+4)}^{2}+3\)

Try it.

\(x=-3{(y-2)}^{2}+3\)

Solution

Try it.

\(x=-2{(y-1)}^{2}+2\)

Try it.

\(x=4{(y+1)}^{2}-4\)

Solution

Try it.

\(x=2{(y+4)}^{2}-2\)

In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.

Try it.

\(x={y}^{2}+4y-5\)

Solution

ⓐ \(x={(y+2)}^{2}-9\)
ⓑ

Try it.

\(x={y}^{2}+2y-3\)

Try it.

\(x=-2{y}^{2}-12y-16\)

Solution

ⓐ \(x=-2{(y+3)}^{2}+2\)
ⓑ

Try it.

\(x=-3{y}^{2}-6y-5\)

Mixed Practice

In the following exercises, match each graph to one of the following equations: ⓐ x² + y² = 64 ⓑ x² + y² = 49
ⓒ (x + 5)² + (y + 2)² = 4 ⓓ (x − 2)² + (y − 3)² = 9 ⓔ y = −x² + 8x − 15 ⓕ y = 6x² + 2x − 1

Solve Applications with Parabolas

Try it.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

Solution

\(y=-\frac{1}{15}{(x-15)}^{2}+15\)

Try it.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

Try it.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

Solution

\(y=-\frac{1}{10}{(x-30)}^{2}+90\)

Try it.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

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Praktyka (40)

Spróbujcie najpierw każdego na papierze. Wykryjcie odpowiedź na sprawdzenie; zweryfikowane można otworzyć w rozwiązującym dla każdego kroku.

1. Graph: \(y=-3{x}^{2}+12x-12.\)

   Przepraszam. Podpowiedź

2. Solve by completing the square: \({x}^{2}-6x+6=0.\)

   Przepraszam. Podpowiedź

   Odkryj odpowiedź

   \(x=3\pm \sqrt{3}\)

3. Write in standard form: \(y=3{x}^{2}-6x+5.\)

   Przepraszam. Podpowiedź

   Odkryj odpowiedź

   \(y=3{\left(x-1\right)}^{2}+2\)

4. Graph \(y=\text{-}{x}^{2}+6x-8\) by using properties.

   Przepraszam. Podpowiedź

   Odkryj odpowiedź

   Since a is \(-1,\) the parabola opens downward.
   To find the axis of symmetry, find \(x=-\frac{b}{2a}.\)
   	The axis of symmetry is \(x=3.\)
   The vertex is on the line \(x=3.\)
   Let \(x=3.\)
   	The vertex is \((3,1).\)
   The y-intercept occurs when \(x=0.\)
   Substitute \(x=0.\)
   Simplify.
   	The y-intercept is \((0,-8).\)
   The point \((0,-8)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,-8).\)	Point symmetric to the y-intercept is \((6,-8).\)
   The x-intercept occurs when \(y=0.\)
   Let \(y=0.\)
   Factor the GCF.
   Factor the trinomial.
   Solve for x.
   	The x-intercepts are \((4,0),(2,0).\)
   Graph the parabola.

5. Graph \(y=\text{-}{x}^{2}+5x-6\) by using properties.

   Przepraszam. Podpowiedź

6. Graph \(y=\text{-}{x}^{2}+8x-12\) by using properties.

   Przepraszam. Podpowiedź

7. Write\(y=3{x}^{2}-6x+5\) in standard form and then use properties of standard form to graph the equation.

   Przepraszam. Podpowiedź

   Odkryj odpowiedź

   Rewrite the function in \(y=a{(x-h)}^{2}+k\) form by completing the square.	\(y=3{x}^{2}-6x+5\)
   	\(y=3({x}^{2}-2x\ )+5\)
   	\(y=3({x}^{2}-2x+1)+5-3\)
   	\(y=3{(x-1)}^{2}+2\)
   Identify the constants a, h, k.	\(a=3\), \(h=1\), \(k=2\)
   Since \(a=3,\) the parabola opens upward.
   The axis of symmetry is \(x=h.\)	The axis of symmetry is \(x=1.\)
   The vertex is \((h,k).\)	The vertex is \((1,2).\)
   Find the y-intercept by substituting \(x=0.\)	\(y=3{(x-1)}^{2}+2\) \(y=3\cdot {0}^{2}-6\cdot 0+5\)
   	\(y=5\)
   	y-intercept \((0,5)\)
   Find the point symmetric to \((0,5)\) across the axis of symmetry.	\((2,5)\)
   Find the x-intercepts.	\(\begin{array}{lll}y & = & 3{(x-1)}^{2}+2 \\ 0 & = & 3{(x-1)}^{2}+2 \\ -2 & = & 3{(x-1)}^{2} \\ -\frac{2}{3} & = & {(x-1)}^{2} \\ \pm \sqrt{-\frac{2}{3}} & = & x-1\end{array}\)
   	The square root of a negative number tells us the solutions are complex numbers. So there are no x-intercepts.
   Graph the parabola.

8. ⓐ Write \(y=2{x}^{2}+4x+5\) in standard form and ⓑ use properties of standard form to graph the equation.

   Przepraszam. Podpowiedź

   Odkryj odpowiedź

   ⓐ \(y=2{(x+1)}^{2}+3\)
   ⓑ

9. ⓐ Write \(y=-2{x}^{2}+8x-7\) in standard form and ⓑ use properties of standard form to graph the equation.

   Przepraszam. Podpowiedź

   Odkryj odpowiedź

   ⓐ \(y=-2{(x-2)}^{2}+1\)
   ⓑ

10. Graph \(x=2{y}^{2}\) by using properties.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    Since \(a=2,\) the parabola opens to the right.
    To find the axis of symmetry, find \(y=-\frac{b}{2a}.\)
    	The axis of symmetry is \(y=0.\)
    The vertex is on the line\(y=0.\)
    Let \(y=0.\)
    	The vertex is \((0,0).\)

    Since the vertex is \((0,0),\) both the x- and y-intercepts are the point \((0,0).\) To graph the parabola we need more points. In this case it is easiest to choose values of y.
    
    We also plot the points symmetric to \((2,1)\) and \((8,2)\) across the y-axis, the points \((2,-1),\)\((8,-2).\)

    Graph the parabola.

11. Graph \(x={y}^{2}\) by using properties.

    Przepraszam. Podpowiedź

12. Graph \(x=\text{-}{y}^{2}\) by using properties.

    Przepraszam. Podpowiedź

13. Graph \(x=\text{-}{y}^{2}+2y+8\) by using properties.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    Since \(a=-1,\) the parabola opens to the left.
    To find the axis of symmetry, find \(y=-\frac{b}{2a}.\)
    	The axis of symmetry is \(y=1.\)
    The vertex is on the line\(y=1.\)
    Let \(y=1.\)
    	The vertex is \((9,1).\)
    The x-intercept occurs when \(y=0.\)
    	The x-intercept is \((8,0).\)
    The point \((8,0)\) is one unit below the line of symmetry. The symmetric point one unit above the line of symmetry is \((8,2)\)	Symmetric point is \((8,2).\)
    The y-intercept occurs when \(x=0.\)
    Substitute \(x=0.\)
    Solve.
    	The y-intercepts are \((0,4)\) and \((0,-2).\)
    Connect the points to graph the parabola.

14. Graph \(x=\text{-}{y}^{2}-4y+12\) by using properties.

    Przepraszam. Podpowiedź

15. Graph \(x=\text{-}{y}^{2}+2y-3\) by using properties.

    Przepraszam. Podpowiedź

16. Graph \(x=2{(y-2)}^{2}+1\) using properties.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    Identify the constants a, h, k.	\(a=2,\)\(h=1,\)\(k=2\)
    Since \(a=2,\) the parabola opens to the right.
    The axis of symmetry is \(y=k.\)	\(\)The axis of symmetry is \(y=2.\)
    The vertex is \((h,k).\)	\(\)The vertex is \((1,2).\)
    Find the x-intercept by substituting \(y=0.\)	\(\begin{array}{lll}x & = & 2{(y-2)}^{2}+1 \\ x & = & 2{(0-2)}^{2}+1 \\ x & = & 9\end{array}\)
    	\(\)The x-intercept is \((9,0).\)
    Find the point symmetric to \((9,0)\) across the axis of symmetry.	\(\ (9,4)\)
    Find the y-intercepts. Let \(x=0.\)	\(\begin{array}{lll}x & = & 2{(y-2)}^{2}+1 \\ 0 & = & 2{(y-2)}^{2}+1 \\ -1 & = & 2{(y-2)}^{2}\end{array}\)
    	A square cannot be negative, so there is no real solution. So there are no y-intercepts.
    Graph the parabola.

17. Graph \(x=3{(y-1)}^{2}+2\) using properties.

    Przepraszam. Podpowiedź

18. Graph \(x=2{(y-3)}^{2}+2\) using properties.

    Przepraszam. Podpowiedź

19. Graph \(x=-4{(y+1)}^{2}+4\) using properties.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    Identify the constants a, h, k.	\(a=-4,\)\(h=4,\)\(k=-1\)
    Since \(a=-4,\) the parabola opens to the left.
    The axis of symmetry is \(y=k.\)	\(\)The axis of symmetry is \(y=-1.\)
    The vertex is \((h,k).\)	\(\)The vertex is \((4,-1).\)
    Find the x-intercept by substituting \(y=0.\)	\(\ \begin{array}{lll}x & = & -4{(y+1)}^{2}+4 \\ x & = & -4{(0+1)}^{2}+4 \\ x & = & 0\end{array}\)
    	\(\)The x-intercept is \((0,0).\)
    Find the point symmetric to \((0,0)\) across the axis of symmetry.	\(\ (0,-2)\)
    Find the y-intercepts.	\(\ x=-4{(y+1)}^{2}+4\)
    Let \(x=0.\)	\(\begin{array}{lll}0 & = & -4{(y+1)}^{2}+4 \\ -4 & = & -4{(y+1)}^{2} \\ 1 & = & {(y+1)}^{2} \\ y+1 & = & \pm 1\end{array}\)
    	\(y=-1+1\ y=-1-1\)
    	\(y=0\ y=-2\)
    	The y-intercepts are \((0,0)\) and \((0,-2).\)
    Graph the parabola.

20. Graph \(x=-4{(y+2)}^{2}+4\) using properties.

    Przepraszam. Podpowiedź

21. Graph \(x=-2{(y+3)}^{2}+2\) using properties.

    Przepraszam. Podpowiedź

22. Write \(x=2{y}^{2}+12y+17\) in standard form and then use the properties of the standard form to graph the equation.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    Rewrite the function in \(x=a{(y-k)}^{2}+h\) form by completing the square.
    Identify the constants a, h, k.	\(a=2,\ h=-1,\ k=-3\)
    Since \(a=2,\) the parabola opens to the right.
    The axis of symmetry is \(y=k.\)	\(\)The axis of symmetry is \(y=-3.\)
    The vertex is \((h,k).\)	\(\)The vertex is \((-1,-3).\)
    Find the x-intercept by substituting \(y=0.\)	\(\ \begin{array}{lll}x & = & 2{(y+3)}^{2}-1 \\ x & = & 2{(0+3)}^{2}-1 \\ x & = & 17\end{array}\)
    	\(\)The x-intercept is \((17,0).\)
    Find the point symmetric to \((17,0)\) across the axis of symmetry.	\(\ (17,-6)\)
    Find the y-intercepts. Let \(x=0.\)	\(\begin{array}{lll}x & = & 2{(y+3)}^{2}-1 \\ 0 & = & 2{(y+3)}^{2}-1 \\ 1 & = & 2{(y+3)}^{2} \\ \frac{1}{2} & = & {(y+3)}^{2} \\ y+3 & = & \pm \sqrt{\frac{1}{2}} \\ y & = & -3\pm \frac{\sqrt{2}}{2}\end{array}\)
    	\(y=-3+\frac{\sqrt{2}}{2}\ y=-3-\frac{\sqrt{2}}{2}\)
    	\(y\approx -2.3\ y\approx -3.7\)
    	The y-intercepts are \((0,-3+\frac{\sqrt{2}}{2}),(0,-3-\frac{\sqrt{2}}{2}).\)
    Graph the parabola.

23. ⓐ Write \(x=3{y}^{2}+6y+7\) in standard form and ⓑ use properties of the standard form to graph the equation.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    ⓐ \(x=3{(y+1)}^{2}+4\)
    ⓑ

24. ⓐ Write \(x=-4{y}^{2}-16y-12\) in standard form and ⓑ use properties of the standard form to graph the equation.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    ⓐ \(x=-4{(y+2)}^{2}+4\)
    ⓑ

25. Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    We will first set up a coordinate system and draw the parabola. The graph will give us the information we need to write the equation of the graph in the standard form\(y=a{(x-h)}^{2}+k.\)
    

    Let the lower left side of the bridge be the origin of the coordinate grid at the point \((0,0).\) Since the base is 20 feet wide the point \((20,0)\) represents the lower right side. The bridge is 10 feet high at the highest point. The highest point is the vertex of the parabola so the y-coordinate of the vertex will be 10. Since the bridge is symmetric, the vertex must fall halfway between the left most point, \((0,0),\) and the rightmost point \((20,0).\) From this we know that the x-coordinate of the vertex will also be 10.
    Identify the vertex, \((h,k).\)	\((h,k)=(10,10)\)
    	\(h=10,\ k=10\)
    Substitute the values into the standard form. The value of a is still unknown. To find the value of a use one of the other points on the parabola.	\(\ \begin{array}{lll}y & = & a{(x-h)}^{2}+k \\ y & = & a{(x-10)}^{2}+10 \\ (x,y) & = & (0,0)\end{array}\)
    Substitute the values of the other point into the equation.	\(\ \begin{array}{lll}y & = & a{(x-10)}^{2}+10 \\ 0 & = & a{(0-10)}^{2}+10\end{array}\)
    Solve for a.	\(\ \begin{array}{lll}0 & = & a{(0-10)}^{2}+10 \\ -10 & = & a{(-10)}^{2} \\ -10 & = & 100a \\ \frac{-10}{100} & = & a \\ a & = & -\frac{1}{10}\end{array}\)
    	\(y=a{(x-10)}^{2}+10\)
    Substitute the value for a into the equation.	\(\ y=-\frac{1}{10}{(x-10)}^{2}+10\)

26. Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    \(y=-\frac{1}{20}{(x-20)}^{2}+20\)

27. Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    \(y=-\frac{1}{5}{(x-5)}^{2}+5\)

28. \(y=\text{-}{x}^{2}+4x-3\)

    Przepraszam. Podpowiedź

29. \(y=\text{-}{x}^{2}+8x-15\)

    Przepraszam. Podpowiedź

30. \(y=6{x}^{2}+2x-1\)

    Przepraszam. Podpowiedź

31. \(y=8{x}^{2}-10x+3\)

    Przepraszam. Podpowiedź

32. \(y=\text{-}{x}^{2}+2x-4\)

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    ⓐ \(y=\text{-}{(x-1)}^{2}-3\)
    ⓑ

33. \(y=2{x}^{2}+4x+6\)

    Przepraszam. Podpowiedź

34. \(y=-2{x}^{2}-4x-5\)

    Przepraszam. Podpowiedź

    Odkryj odpowiedź

    ⓐ \(y=-2{(x+1)}^{2}-3\)
    ⓑ

35. \(y=3{x}^{2}-12x+7\)

    Przepraszam. Podpowiedź

36. \(x=-2{y}^{2}\)

    Przepraszam. Podpowiedź

37. \(x=3{y}^{2}\)

    Przepraszam. Podpowiedź

38. \(x=4{y}^{2}\)

    Przepraszam. Podpowiedź

39. \(x=-4{y}^{2}\)

    Przepraszam. Podpowiedź

40. \(x=\text{-}{y}^{2}-2y+3\)

    Przepraszam. Podpowiedź

Zachowaj swoją pracę.

Darmowe konto dodaje notatki na każdej lekcji, zapis tego, co zakończyłeś, rozwiązane problemy w jednym miejscu, a także korepetytor, którego możesz zapytać o tę stronę. Matematyka sama jest otwarta dla wszystkich, zapisana lub nie.

Podpisz Zalogowanie

Symbole używane tutaj

Dotknij dowolny symbol pełnej definicji, obrazu i co oznacza każda litera.

|x|absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.\binom{n}{k}binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).\pipi
Ratio of a circle's circumference to its diameter, 3.14159…eEuler's number
2.71828…, the base whose exponential is its own derivative.iimaginary unit
i² = −1.\thetatheta
The usual name for an angle.\log_b xlogarithm, natural log
The exponent b must be raised to for x; ln uses base e.\mathbf{v}vector
A quantity with magnitude and direction; a column of numbers.

Jak to zrobić?: Parabolas

1. Graph vertical parabolas
2. Graph horizontal parabolas
3. Solve applications with parabolas
4. Determine whether the parabola opens upward or downward.
5. Find the axis of symmetry.
6. Find the vertex.
7. Find the
8. Find the

Pytania, które ludzie zadają

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Części tej strony są dostosowane z OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Tu są zgłośliwe i ponownie wyjaśnione; błędy są nasze.

Więcej w Precalculus

Complex numbersPolynomial functionsRational functionsSequences and seriesThe binomial theoremConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates