Lesson 9.7: Graph Quadratic Functions Using Transformations

New Concept

Learn to graph any quadratic function by starting with the basic parabola, $f(x) = x^2$, and applying transformations. We'll shift it vertically and horizontally, and stretch or compress it to match its vertex form, $f(x) = a(x-h)^2+k$.

What’s next

Next, you’ll see exactly how the parameters $a$, $h$, and $k$ shift and stretch the parabola through interactive examples and practice problems.

Property

The graph of $f(x) = x^2 + k$ shifts the graph of $f(x) = x^2$ vertically $k$ units.

• If $k > 0$, shift the parabola vertically up $k$ units.
• If $k < 0$, shift the parabola vertically down $|k|$ units.

Examples

• To graph $f(x) = x^2 + 4$, you start with the graph of $f(x) = x^2$ and shift it vertically up 4 units because $k = 4$.

• To graph $f(x) = x^2 - 5$, you start with the graph of $f(x) = x^2$ and shift it vertically down 5 units because $k = -5$.

Property

The graph of $f(x) = (x - h)^2$ shifts the graph of $f(x) = x^2$ horizontally $h$ units.

• If $h > 0$, shift the parabola horizontally right $h$ units.
• If $h < 0$, shift the parabola horizontally left $|h|$ units.

Examples

• To graph $f(x) = (x - 3)^2$, you shift the graph of $f(x) = x^2$ to the right 3 units. The vertex moves from $(0, 0)$ to $(3, 0)$.

• To graph $f(x) = (x + 4)^2$, you rewrite it as $f(x) = (x - (-4))^2$. This means you shift the graph of $f(x) = x^2$ to the left 4 units.

Property

The coefficient $a$ in the function $f(x) = ax^2$ affects the graph of $f(x) = x^2$ by stretching or compressing it.

• If $0 < |a| < 1$, the graph of $f(x) = ax^2$ will be “wider” than the graph of $f(x) = x^2$.
• If $|a| > 1$, the graph of $f(x) = ax^2$ will be “skinnier” than the graph of $f(x) = x^2$.

Examples

• The graph of $f(x) = 4x^2$ is skinnier than $f(x) = x^2$. For any given x-value, the y-value is multiplied by 4, stretching the parabola vertically.

• The graph of $f(x) = \frac{1}{3}x^2$ is wider than $f(x) = x^2$. The y-values are compressed to one-third of their original height, making the parabola open more broadly.

Property

How To Graph a quadratic function using transformations
Step 1. Rewrite the function in $f(x) = a(x - h)^2 + k$ form by completing the square. This is also known as the vertex form.
Step 2. Graph the function by applying transformations to the basic graph of $f(x) = x^2$. Apply stretch/compression ($a$), then horizontal shift ($h$), then vertical shift ($k$).

Examples

• To graph $f(x) = x^2 + 4x + 1$, complete the square to get $f(x) = (x+2)^2 - 3$. This is the graph of $x^2$ shifted left 2 units and down 3 units.

• To graph $f(x) = 3x^2 - 6x + 5$, complete the square to get $f(x) = 3(x-1)^2 + 2$. This is $x^2$ stretched by 3, shifted right 1, and up 2.

Property

How To Graph a quadratic function in the form $f(x) = a(x - h)^2 + k$ using properties

1. Determine if the parabola opens upward ($a > 0$) or downward ($a < 0$).
2. Find the axis of symmetry, $x = h$.
3. Find the vertex, $(h, k)$.
4. Find the $y$-intercept by calculating $f(0)$.
5. Find the $x$-intercepts by solving $f(x) = 0$.
6. Graph the parabola using these key points.

Examples

• For $f(x) = 3(x-2)^2 + 5$, the parabola opens upward ($a=3$), the vertex is $(2, 5)$, and the axis of symmetry is $x=2$.

• For $f(x) = -(x+1)^2 - 4$, the parabola opens downward ($a=-1$), the vertex is $(-1, -4)$, and the axis of symmetry is $x=-1$.

Property

To find a quadratic function from its graph, start with the vertex form $f(x) = a(x - h)^2 + k$.

1. Identify the vertex $(h, k)$ from the graph and substitute these values into the equation.
2. Identify another point on the graph, such as the $y$-intercept.
3. Substitute the coordinates of this point for $x$ and $f(x)$, then solve for the coefficient $a$.
4. Write the final function using the values of $a$, $h$, and $k$.

Examples

• A parabola has a vertex at $(2, 3)$ and passes through $(0, 7)$. Start with $f(x) = a(x-2)^2+3$. Use $(0, 7)$ to find $a$: $7 = a(0-2)^2+3$, so $4=4a$ and $a=1$. The function is $f(x) = (x-2)^2+3$.

• A parabola has a vertex at $(-1, -5)$ and a y-intercept at $(0, -3)$. Start with $f(x) = a(x+1)^2-5$. Use $(0, -3)$ to find $a$: $-3 = a(0+1)^2-5$, so $2=a$. The function is $f(x) = 2(x+1)^2-5$.