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10.2: Vertical Shifts of Quadratic Functions

Difficulty Level: Basic Created by: CK-12
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Suppose the marketing department of a company is designing a new logo that includes a parabola. They've drawn the logo on a piece of graph paper, but they've decided that they want to change the position of the parabola by moving it 3 units down. If the original equation of the parabola was \begin{align*}y=2x^2\end{align*}, what would be the new equation of the parabola after they perform the vertical shift? How do you know? In this Concept, you'll learn about vertical shifts of quadratic functions so that you can analyze a situation like this one.

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Multimedia Link: For more information regarding stopping distance, watch this: CK-12 Basic Algebra: Algebra Applications: Quadratic Functions

- YouTube video.


Compare the five parabolas. What do you notice?

The five different parabolas are congruent with different \begin{align*}y-\end{align*}intercepts. Each parabola has an equation of the form \begin{align*}y=ax^2+c\end{align*}, where \begin{align*}a=1\end{align*} and \begin{align*}c=y-\end{align*}intercept. In general, the value of \begin{align*}c\end{align*} will tell you where the parabola will intersect the \begin{align*}y-\end{align*}axis.

The equation \begin{align*}y=ax^2+c\end{align*} is a parabola with a \begin{align*}y-\end{align*}intercept of \begin{align*}(0, c)\end{align*}.

The vertical movement along a parabola’s line of symmetry is called a vertical shift.

Example A

Determine the direction, shape, and \begin{align*}y-\end{align*}intercept of the parabola formed by \begin{align*}y=\frac{3}{2} x^2-4\end{align*}.

Solution: The value of \begin{align*}a\end{align*} in the quadratic equation is \begin{align*}\frac{3}{2}\end{align*}.

  • Because \begin{align*}a\end{align*} is positive, the parabola opens upward.
  • Because \begin{align*}a\end{align*} is greater than 1, the parabola is narrow about its line of symmetry.
  • The value of \begin{align*}c\end{align*} is –4, so the \begin{align*}y-\end{align*}intercept is (0, –4).

Projectiles are often described by quadratic equations. When an object is dropped from a tall building or cliff, it does not travel at a constant speed. The longer it travels, the faster it goes. Galileo described this relationship between distance fallen and time. It is known as his kinematical law. It states the “distance traveled varies directly with the square of time.” As an algebraic equation, this law is:


Use this information to graph the distance an object travels during the first six seconds.

\begin{align*}t\end{align*} \begin{align*}d\end{align*}
0 0
1 16
2 64
3 144
4 256
5 400
6 576

The parabola opens upward, and its vertex is located at the origin. Since \begin{align*}a>1\end{align*}, the graph is narrow about its line of symmetry. However, because the values of the dependent variable, \begin{align*}d\end{align*}, are very large, the graph is misleading.

Example B

Anne is playing golf. On the fourth tee, she hits a slow shot down the level fairway. The ball follows a parabolic path described by the equation, \begin{align*}y=x-0.04x^2\end{align*}, where \begin{align*}x=\end{align*} distance in feet from the tee and \begin{align*}y=\end{align*} height of the golf ball, in feet.

Describe the shape of this parabola. What is its \begin{align*}y-\end{align*}intercept?

Solution: The value of \begin{align*}a\end{align*} in the quadratic equation is –0.04.

  • Because \begin{align*}a\end{align*} is negative, the parabola opens downward.
  • Because \begin{align*}a\end{align*} is between –1 and 1, the parabola is wide about its line of symmetry.
  • The value of \begin{align*}c\end{align*} is 0, so the \begin{align*}y-\end{align*}intercept is (0, 0).

The distance it takes a car to stop (in feet) given its speed (in miles per hour) is given by the function \begin{align*}d(s)=\frac{1}{20} \ s^2+s\end{align*}. This equation is in standard form: \begin{align*}f(x)=ax^2+bx+c\end{align*}, where \begin{align*}a=\frac{1}{20}, b=1\end{align*}, and \begin{align*}c=0\end{align*}.

Graph the function by making a table of speed values.

\begin{align*}s\end{align*} \begin{align*}d\end{align*}
0 0
10 15
20 40
30 75
40 120
50 175
60 240
  • The parabola opens upward with a vertex at (0, 0).
  • The line of symmetry is \begin{align*}x=0\end{align*}.
  • The parabola is wide about its line of symmetry.

Using the function to find the stopping distance of a car travelling 65 miles per hour yields:

\begin{align*}d(65)=\frac{1}{20} (65)^2+65=276.25 \ feet\end{align*}

The Affect of Vertical Shifting on the \begin{align*}x-\end{align*}Intercepts

Consider the graphs of quadratic functions in the beginning of this Concept:

The graph of \begin{align*}y=x^2+2\end{align*} has no \begin{align*}x\end{align*}-intercepts; as we shift it down 1, it still has no \begin{align*}x\end{align*}-intercepts. However, shifting down to \begin{align*}y=x^2\end{align*}, this parabola has one \begin{align*}x\end{align*}-intercept, at the vertex. Notice that shifting down one more to \begin{align*}y=x^2-1\end{align*}, there are two \begin{align*}x\end{align*}-intercepts.

Example C

Find the \begin{align*}x\end{align*}-intercepts of \begin{align*}y=x^2-1\end{align*} by factoring.


We can see that the \begin{align*}x\end{align*}-intercepts are -1 and 1 from the graph above. Use this to check your factoring.

To factor \begin{align*}y=x^2-1\end{align*}, notice that it fits the pattern \begin{align*}y=a^2-b^2=(a-b)(a+b)\end{align*}. Thus:


Since \begin{align*}y=0\end{align*} at the \begin{align*}x\end{align*}-intercepts, solve:

\begin{align*}0=y=(x-1)(x+1)\Rightarrow (x-1)=0 \text{ or }(x+1)=0 \Rightarrow x=1 \text{ or }x=-1 .\end{align*}

So, the \begin{align*}x\end{align*}-intercepts are -1 and 1.

Guided Practice

Determine the direction, shape, \begin{align*}y\end{align*}-intercept, and \begin{align*}x\end{align*}-intercepts of \begin{align*}y=-x^2-3x+18\end{align*}.


Since \begin{align*}a=-1\end{align*}, the direction is down. The shape is neither wide nor narrow about the axis of symmetry. The \begin{align*}y\end{align*}-intercept can be found by substituting in \begin{align*}x=0\end{align*}:


So the \begin{align*}y\end{align*}-intercept is \begin{align*}(0,18)\end{align*}. Notice 18 is also the value of \begin{align*}c\end{align*}. This is always true for quadratic functions.

To find the \begin{align*}x\end{align*}-intercepts, factor:


This means that \begin{align*}y=0\end{align*} when

\begin{align*}x+6=0 \text{ or } x-3=0 \Rightarrow x=-6 \text{ or } x=3.\end{align*}

Thus, the \begin{align*}x\end{align*}-intercepts are -6 and 3.


Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Graphs of Quadratic Functions (16:05)

  1. Using the parabola below, identify the following:
    1. Vertex
    2. \begin{align*}y-\end{align*}intercept
    3. \begin{align*}x-\end{align*}intercepts
    4. Domain
    5. Range
    6. Line of symmetry
    7. Is \begin{align*}a\end{align*} positive or negative?
    8. Is \begin{align*}a\end{align*} \begin{align*}-1<a<1\end{align*} or \begin{align*}a<-1\end{align*} or \begin{align*}a>1\end{align*}?

  1. Use the stopping distance function from the Concept to find:
    1. \begin{align*}d(45)\end{align*}
    2. What speed has a stopping distance of about 96 feet?
  1. Using Galileo’s Law from the Concept, find:
    1. The distance an object has fallen at 3.5 seconds
    2. The total distance the object has fallen in 3.5 seconds

Which has a more positive \begin{align*}y-\end{align*}intercept?

  1. \begin{align*}y=x^2\end{align*} or \begin{align*}y=4x^2\end{align*}
  2. \begin{align*}y=2x^2+4\end{align*} or \begin{align*}y=\frac{1}{2} x^2+4\end{align*}
  3. \begin{align*}y=-2x^2-2\end{align*} or \begin{align*}y=-x^2-2\end{align*}

Identify the vertex and \begin{align*}y-\end{align*}intercept. Is the vertex a maximum or a minimum?

  1. \begin{align*}y=x^2-2x-8\end{align*}
  2. \begin{align*}y=-x^2+10x-21\end{align*}
  3. \begin{align*}y=2x^2+6x+4\end{align*}

Which equation has a larger vertex?

  1. \begin{align*}y=x^2\end{align*} or \begin{align*}y=4x^2\end{align*}
  2. \begin{align*}y=-2x^2\end{align*} or \begin{align*}y=-2x^2 -2\end{align*}
  3. \begin{align*}y=3x^2-3\end{align*} or \begin{align*}y=3x^2-6\end{align*}
  1. Nadia is throwing a ball to Peter. Peter does not catch the ball and it hits the ground. The graph shows the path of the ball as it flies through the air. The equation that describes the path of the ball is \begin{align*}y=4+2x-0.16x^2\end{align*}. Here, \begin{align*}y\end{align*} is the height of the ball and \begin{align*}x\end{align*} is the horizontal distance from Nadia. Both distances are measured in feet. How far from Nadia does the ball hit the ground? At what distance, \begin{align*}x\end{align*}, from Nadia, does the ball attain its maximum height? What is the maximum height?
  2. Peter wants to enclose a vegetable patch with 120 feet of fencing. He wants to put the vegetable patch against an existing wall, so he needs fence for only three of the sides. The equation for the area is given by \begin{align*}a=120 x-x^2\end{align*}. From the graph, find what dimensions of the rectangle would give him the greatest area.

Mixed Review

  1. Factor \begin{align*}6u^2 v-11u^2 v^2-10u^2 v^3\end{align*} using its GCF.
  2. Factor into primes: \begin{align*}3x^2+11x+10\end{align*}.
  3. Simplify \begin{align*}- \frac{1}{9} (63) \left(- \frac{3}{7} \right)\end{align*}.
  4. Solve for \begin{align*}b: |b+2|=9\end{align*}.
  5. Simplify \begin{align*}(4x^3 y^2 z)^3\end{align*}.
  6. What is the slope and \begin{align*}y-\end{align*}intercept of \begin{align*}7x+4y=9?\end{align*}

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vertical shifts The vertical movement along a parabola’s line of symmetry is called a vertical shift.

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Difficulty Level:
8 , 9
Date Created:
Feb 24, 2012
Last Modified:
Apr 11, 2016
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