9.7 Graph Quadratic Functions Using Transformations

Graphing Quadratic Functions Using Transformations

Every quadratic function produces a parabola, and every parabola is just a transformed version of the parent function $y = x^2$. Instead of plotting point after point, you can use transformations to shift, stretch, compress, or flip that parent parabola into the correct graph. The tool that makes this work is vertex form: $f(x) = a(x - h)^2 + k$.

In this form, each parameter controls a specific transformation:

• $a$ controls the direction (up or down) and how narrow or wide the parabola is
• $h$ controls the horizontal shift (left or right)
• $k$ controls the vertical shift (up or down)

The vertex of the parabola sits at the point $(h, k)$, which gives you an immediate anchor for your graph.

Vertical Shifts in Quadratic Functions

A vertical shift moves the entire parabola up or down without changing its shape. The parameter $k$ in vertex form controls this.

• If $k > 0$, the graph shifts up by $k$ units
• If $k < 0$, the graph shifts down by $|k|$ units

To graph with a vertical shift:

1. Start with the parent function $y = x^2$, which has its vertex at $(0, 0)$
2. Move the vertex up or down by $k$ units to get the new vertex at $(h, k)$
3. Determine the direction the parabola opens based on the sign of $a$
4. Sketch the parabola from the new vertex, keeping the same shape as the parent

Examples:

• $f(x) = (x - 2)^2 + 3$ has vertex at $(2, 3)$. The $+3$ shifts the parabola up 3 units from where it would otherwise sit.
• $g(x) = -(x + 1)^2 - 4$ has vertex at $(-1, -4)$. The $-4$ shifts the parabola down 4 units. The negative $a$ also flips it to open downward.

Horizontal Shifts of Quadratic Graphs

Horizontal shifts move the parabola left or right. The parameter $h$ controls this, but watch the sign carefully: the vertex form has $(x - h)$, so the shift works opposite to what you might expect.

• $(x - 3)^2$ means $h = 3$, so the graph shifts right 3 units
• $(x + 2)^2$ means $h = -2$, so the graph shifts left 2 units

The sign flip trips up a lot of students. Think of it this way: $(x + 2)^2$ is really $(x - (-2))^2$, so $h = -2$.

To graph with a horizontal shift:

1. Identify the vertex $(h, k)$ by reading $h$ and $k$ from vertex form
2. Plot the vertex on the coordinate plane
3. Determine the direction the parabola opens based on the sign of $a$
4. Sketch the parabola from the new vertex, shifted horizontally from the parent function

Examples:

• $f(x) = (x - 3)^2$ shifts the parent parabola right 3 units. The vertex moves to $(3, 0)$.
• $g(x) = (x + 2)^2$ shifts the parent parabola left 2 units. The vertex moves to $(-2, 0)$.

Stretching vs. Compressing Quadratic Functions

The value of $a$ controls how wide or narrow the parabola is compared to the parent function $y = x^2$ (where $a = 1$).

• If $|a| > 1$, the parabola is narrower (a vertical stretch that pulls the graph away from the x-axis)
• If $0 < |a| < 1$, the parabola is wider (a vertical compression that pushes the graph toward the x-axis)

A common point of confusion: when $|a| > 1$, the y-values grow faster, so the parabola looks narrower even though the transformation is called a vertical stretch. Think about it as each y-value being multiplied by $a$. For the parent function at $x = 1$, $y = 1$. But for $f(x) = 2x^2$, at $x = 1$, $y = 2$. The points are higher (farther from the x-axis), which makes the curve steeper and narrower.

Examples:

• $f(x) = 2x^2$: each y-value is doubled compared to $y = x^2$, making the parabola narrower
• $g(x) = \frac{1}{3}x^2$: each y-value is one-third of the parent, making the parabola wider

If $a$ is negative, the parabola also reflects (flips) across the x-axis, opening downward instead of upward.

Multiple Transformations of Quadratics

Most quadratic functions involve several transformations at once. Here's how to handle them systematically:

1. Identify $a$, $h$, and $k$ from the vertex form $f(x) = a(x - h)^2 + k$

2. Plot the vertex at $(h, k)$

3. Determine direction: if $a > 0$ the parabola opens up; if $a < 0$ it opens down

4. Determine width: compare $|a|$ to 1 to decide if the parabola is narrower or wider than the parent

5. Plot a few additional points by choosing x-values near the vertex and calculating their y-values

6. Sketch the parabola through your points

Example: Graph $f(x) = -2(x + 1)^2 + 3$

1. Rewrite as $f(x) = -2(x - (-1))^2 + 3$, so $a = -2$, $h = -1$, $k = 3$

2. Vertex is at $(-1, 3)$

3. Since $a = -2 < 0$, the parabola opens downward

4. Since $|a| = 2 > 1$, the parabola is narrower than the parent

5. Check a point: at $x = 0$, $f(0) = -2(0 + 1)^2 + 3 = -2(1) + 3 = 1$. So $(0, 1)$ is on the graph. By symmetry, $(-2, 1)$ is also on the graph.

6. Sketch a narrow, downward-opening parabola through these points

Equations from Quadratic Graphs

You can also work backward: given a graph, find the equation in vertex form.

1. Read the vertex $(h, k)$ directly from the graph

2. Determine the sign of $a$: positive if the parabola opens upward, negative if downward

3. Find the value of $|a|$ by using another point on the graph. Pick a point $(x, y)$ that you can read clearly, then solve $y = a(x - h)^2 + k$ for $a$

4. Write the equation: $f(x) = a(x - h)^2 + k$

Example: A parabola has its vertex at $(2, -3)$, opens upward, and passes through the point $(4, -1)$.

1. $h = 2$, $k = -3$

2. Opens upward, so $a > 0$

3. Plug in $(4, -1)$: $-1 = a(4 - 2)^2 + (-3)$, which gives $-1 = 4a - 3$, so $4a = 2$, and $a = \frac{1}{2}$

4. The equation is $f(x) = \frac{1}{2}(x - 2)^2 - 3$

Using an actual point from the graph to solve for $a$ is more reliable than just estimating whether the parabola looks wider or narrower.

Additional Properties of Quadratic Functions

Domain and Range:

• The domain of any quadratic function is all real numbers: $(-\infty, \infty)$
• The range depends on the vertex and direction of opening:
  • Opens upward ($a > 0$): range is $[k, \infty)$
  • Opens downward ($a < 0$): range is $(-\infty, k]$

Concavity:

• $a > 0$: concave up (opens upward, vertex is a minimum)
• $a < 0$: concave down (opens downward, vertex is a maximum)

Zeros (x-intercepts):

The x-intercepts are where $f(x) = 0$. You can find them by solving $a(x - h)^2 + k = 0$. The number of x-intercepts depends on the relationship between the vertex and the direction of opening:

• If the parabola opens upward and $k < 0$, there are two x-intercepts
• If the parabola opens upward and $k = 0$, there is one x-intercept (the vertex touches the x-axis)
• If the parabola opens upward and $k > 0$, there are no x-intercepts

The same logic applies in reverse for downward-opening parabolas. This connects to the discriminant $b^2 - 4ac$ when the equation is in standard form $ax^2 + bx + c$: positive discriminant means two zeros, zero discriminant means one, and negative discriminant means none.