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Study Guides > College Algebra CoRequisite Course

Transformations of Quadratic Functions

Learning Outcomes

  • Graph vertical and horizontal shifts of quadratic functions
  • Graph vertical compressions and stretches of quadratic functions
  • Write the equation of a transformed quadratic function using the vertex form
  • Identify the vertex and axis of symmetry for a given quadratic function in vertex form

The standard form of a quadratic function presents the function in the form

f(x)=a(xh)2+kf\left(x\right)=a{\left(x-h\right)}^{2}+k

where (h, k)\left(h,\text{ }k\right) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.

The standard form is useful for determining how the graph is transformed from the graph of y=x2y={x}^{2}. The figure below is the graph of this basic function. Graph of y=x^2.  

Shift Up and Down by Changing the Value of kk

You can represent a vertical (up, down) shift of the graph of f(x)=x2f(x)=x^2 by adding or subtracting a constant, kk.

f(x)=x2+kf(x)=x^2 + k

 If k>0k>0, the graph shifts upward, whereas if k<0k<0, the graph shifts downward.

Example

Using an online graphing calculator, plot the function f(x)=x2+kf(x)=x^2+k. Now change the kk value to shift the graph down 4 units, then up 4 units.

Answer: The equation for the graph of f(x)=x2f(x)=x^2 that has been shifted up 4 units is f(x)=x2+4f(x)=x^2+4 The equation for the graph of f(x)=x2f(x)=x^2 that has been shifted down 4 units is f(x)=x24f(x)=x^2-4

Shift left and right by changing the value of hh

You can represent a horizontal (left, right) shift of the graph of f(x)=x2f(x)=x^2 by adding or subtracting a constant, hh, to the variable xx, before squaring.

f(x)=(xh)2f(x)=(x-h)^2

If h>0h>0, the graph shifts toward the right and if h<0h<0, the graph shifts to the left.

tip for success

Remember that the negative sign inside the argument of the vertex form of a parabola (in the parentheses with the variable xx ) is part of the formula f(x)=(xh)2+kf(x)=(x-h)^2 +k. If h>0h>0, we have f(x)=(xh)2+kf(x)=(x-h)^2 +k. You'll see the negative sign, but the graph will shift right. If  h<0h<0, we have f(x)=(x(h))2+kf(x)=(x+h)2+kf(x)=(x-(-h))^2 +k \rightarrow f(x)=(x+h)^2+k. You'll see the positive sign, but the graph will shift left.

Example

Using an online graphing calculator, plot the function f(x)=(xh)2f(x)=(x-h)^2. Now change the hh value to shift the graph 2 units to the right, then 2 units to the left.

Answer: The equation for the graph of f(x)=x2f(x)=x^2 that has been shifted right 2 units is

f(x)=(x2)2f(x)=(x-2)^2

The equation for the graph of f(x)=2f(x)=^2 that has been shifted left 2 units is

f(x)=(x+2)2f(x)=(x+2)^2

Stretch or compress by changing the value of aa.

You can represent a stretch or compression (narrowing, widening) of the graph of f(x)=x2f(x)=x^2 by multiplying the squared variable by a constant, aa.

f(x)=ax2f(x)=ax^2

The magnitude of aa indicates the stretch of the graph. If a>1|a|>1, the point associated with a particular xx-value shifts farther from the xx-axis, so the graph appears to become narrower, and there is a vertical stretch. But if a<1|a|<1, the point associated with a particular xx-value shifts closer to the xx-axis, so the graph appears to become wider, but in fact there is a vertical compression.

Example

Using an online graphing calculator plot the function f(x)=ax2f(x)=ax^2. Now adjust the aa value to create a graph that has been compressed vertically by a factor of 12\frac{1}{2} and another that has been vertically stretched by a factor of 3. What are the equations of the two graphs?

Answer: The equation for the graph of f(x)=x2f(x)=x^2 that has been compressed vertically by a factor of 12\frac{1}{2} is

f(x)=12x2f(x)=\frac{1}{2}x^2

The equation for the graph of f(x)=x2f(x)=x^2 that has been vertically stretched by a factor of 3 is

f(x)=3x2f(x)=3x^2

The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.
\begin{align}&a{\left(x-h\right)}^{2}+k=a{x}^{2}+bx+c\\ &a{x}^{2}-2ahx+\left(a{h}^{2}+k\right)=a{x}^{2}+bx+c \end{align}
For the two sides to be equal, the corresponding coefficients must be equal. In particular, the coefficients of xx must be equal.
2ah=b, so h=b2a-2ah=b,\text{ so }h=-\dfrac{b}{2a}.

This is the xx coordinate of the vertexr and x=b2ax=-\dfrac{b}{2a} is the axis of symmetry we defined earlier. Setting the constant terms equal gives us:

\begin{align}a{h}^{2}+k&=c \\[2mm] k&=c-a{h}^{2} \\ &=c-a-{\left(\dfrac{b}{2a}\right)}^{2} \\ &=c-\dfrac{{b}^{2}}{4a} \end{align}
In practice, though, it is usually easier to remember that hh is the output value of the function when the input is hh, so f(h)=f(b2a)=kf\left(h\right)=f\left(-\dfrac{b}{2a}\right)=k.

Example

Using an online graphing calculator plot the function f(x) = a(xh)2+kf\left(x\right)\ =\ a\left(x-h\right)^2+k. Now adjust the variables a,h,ka,h,k to define two quadratic functions whose axis of symmetry is x=3x=-3, and whose vertex is (3,2)(-3, 2). How many potential values are there for hh in this scenario? How about kk? How about aa

Answer: There is only one (h,k)(h,k) pair that will satisfy these conditions, (3,2)(-3,2).  The value of aa does not affect the line of symmetry or the vertex of a quadratic graph, so aa can take on any value.

 Challenge Problem

Define a function whose axis of symmetry is x=3x = -3, and whose vertex is (3,2)(-3,2) and has an average rate of change of 2 on the interval [2,0][-2,0]. Use the online graphing calculator where you plotted f(x) = a(xh)2+kf\left(x\right)\ =\ a\left(x-h\right)^2+k to help with this.

Try It

A coordinate grid has been superimposed over the quadratic path of a basketball in the picture below. Find an equation for the path of the ball. Does the shooter make the basket?
Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes. (credit: modification of work by Dan Meyer)

Answer: The path passes through the origin and has vertex at (4, 7)\left(-4,\text{ }7\right), so (h)x=716(x+4)2+7\left(h\right)x=-\frac{7}{16}{\left(x+4\right)}^{2}+7. To make the shot, h(7.5)h\left(-7.5\right) would need to be about 4 but h(7.5)1.64h\left(-7.5\right)\approx 1.64; he doesn’t make it.

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