Graphing quadratic functions in intercept form worksheet with six equations and blank graphs for plotting.

Let's solve each problem step by step using the intercept form of a quadratic function:

The intercept form is:
$$
y = a(x - p)(x - q)
$$
where:
- $p$ and $q$ are the x-intercepts (roots),
- The axis of symmetry is the average of the x-intercepts:
$$
x = \frac{p + q}{2}
$$
- The vertex lies on the axis of symmetry. Plug that $x$-value into the equation to find the $y$-coordinate.
- The y-intercept is found by plugging $x = 0$ into the equation.

---

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

#### x-intercepts:
Set $y = 0$:
$$
\frac{1}{2}(x + 4)(x - 2) = 0 \Rightarrow x = -4, x = 2
$$
So, x-intercepts: $(-4, 0)$, $(2, 0)$

#### Axis of Symmetry:
$$
x = \frac{-4 + 2}{2} = \frac{-2}{2} = -1
$$

#### Vertex:
Plug $x = -1$ into the equation:
$$
y = \frac{1}{2}(-1 + 4)(-1 - 2) = \frac{1}{2}(3)(-3) = \frac{1}{2}(-9) = -4.5
$$
Vertex: $(-1, -4.5)$

#### y-intercept:
Set $x = 0$:
$$
y = \frac{1}{2}(0 + 4)(0 - 2) = \frac{1}{2}(4)(-2) = \frac{1}{2}(-8) = -4
$$
y-intercept: $(0, -4)$

#### One more point:
Try $x = 1$:
$$
y = \frac{1}{2}(1 + 4)(1 - 2) = \frac{1}{2}(5)(-1) = -2.5 \Rightarrow (1, -2.5)
$$

✔ Answer:
- x-intercepts: $(-4, 0), (2, 0)$
- Axis of symmetry: $x = -1$
- Vertex: $(-1, -4.5)$
- y-intercept: $(0, -4)$
- Extra point: $(1, -2.5)$

---

2) $ y = -\frac{1}{2}(x - 8) $

Wait! This seems incomplete — it says $y = -\frac{1}{2}(x - 8)$, but that’s linear, not quadratic. But looking at the context, likely a typo. Probably meant:
$$
y = -\frac{1}{2}(x - 8)(x - \text{?})
$$
But in the image, it shows only one factor: $y = -\frac{1}{2}(x - 8)$

Wait — actually, let's check the original image description:
"2) $ y = -\frac{1}{2}(x - 8) $" — this is not a quadratic unless there's a second factor.

But the format suggests it should be two factors. Let's assume it's a typo and perhaps it's supposed to be:
$$
y = -\frac{1}{2}(x - 8)(x - ?)
$$
But since no second factor is given, maybe it's missing.

Wait — rechecking: The worksheet says "Practice Worksheet: Graphing Quadratic Functions in Intercept Form", so all equations must be quadratic.

Possibility: It's $ y = -\frac{1}{2}(x - 8)(x - a) $, but only one factor is shown.

Alternatively, maybe it's $ y = -\frac{1}{2}(x - 8)^2 $? But that would be vertex form.

Wait — actually, looking again:
It says:
2) $ y = -\frac{1}{2}(x - 8) $ — this is linear, not quadratic.

This is likely a typo. Perhaps it was meant to be:
$$
y = -\frac{1}{2}(x - 8)(x - b)
$$
But without a second factor, we can't proceed.

Wait — perhaps it's $ y = -\frac{1}{2}(x - 8)(x) $? That would make sense.

Looking at standard problems, maybe it's $ y = -\frac{1}{2}(x)(x - 8) $

Let me assume that it's:
$$
y = -\frac{1}{2}x(x - 8)
$$
That makes sense for intercept form.

So:
$$
y = -\frac{1}{2}(x)(x - 8)
$$

Now solve:

#### x-intercepts:
Set $y = 0$:
$$
-\frac{1}{2}x(x - 8) = 0 \Rightarrow x = 0, x = 8
$$
x-intercepts: $(0, 0)$, $(8, 0)$

#### Axis of symmetry:
$$
x = \frac{0 + 8}{2} = 4
$$

#### Vertex:
Plug $x = 4$:
$$
y = -\frac{1}{2}(4)(4 - 8) = -\frac{1}{2}(4)(-4) = -\frac{1}{2}(-16) = 8
$$
Vertex: $(4, 8)$

#### y-intercept:
Set $x = 0$:
$$
y = -\frac{1}{2}(0)(-8) = 0 \Rightarrow (0, 0)
$$

#### One more point:
Try $x = 2$:
$$
y = -\frac{1}{2}(2)(2 - 8) = -\frac{1}{2}(2)(-6) = -\frac{1}{2}(-12) = 6 \Rightarrow (2, 6)
$$

✔ Answer:
- x-intercepts: $(0, 0), (8, 0)$
- Axis of symmetry: $x = 4$
- Vertex: $(4, 8)$
- y-intercept: $(0, 0)$
- Extra point: $(2, 6)$

---

3) $ y = (x + 2)(x - 2) $

This is difference of squares: $y = x^2 - 4$

#### x-intercepts:
Set $y = 0$:
$$
(x + 2)(x - 2) = 0 \Rightarrow x = -2, x = 2
$$
x-intercepts: $(-2, 0), (2, 0)$

#### Axis of symmetry:
$$
x = \frac{-2 + 2}{2} = 0
$$

#### Vertex:
Plug $x = 0$:
$$
y = (0 + 2)(0 - 2) = (2)(-2) = -4
$$
Vertex: $(0, -4)$

#### y-intercept:
Set $x = 0$: same as above → $(0, -4)$

#### One more point:
Try $x = 1$:
$$
y = (1 + 2)(1 - 2) = (3)(-1) = -3 \Rightarrow (1, -3)
$$

✔ Answer:
- x-intercepts: $(-2, 0), (2, 0)$
- Axis of symmetry: $x = 0$
- Vertex: $(0, -4)$
- y-intercept: $(0, -4)$
- Extra point: $(1, -3)$

---

4) $ y = -\frac{1}{3}(x + 1)(x - 5) $

#### x-intercepts:
Set $y = 0$:
$$
x = -1, x = 5
$$
x-intercepts: $(-1, 0), (5, 0)$

#### Axis of symmetry:
$$
x = \frac{-1 + 5}{2} = \frac{4}{2} = 2
$$

#### Vertex:
Plug $x = 2$:
$$
y = -\frac{1}{3}(2 + 1)(2 - 5) = -\frac{1}{3}(3)(-3) = -\frac{1}{3}(-9) = 3
$$
Vertex: $(2, 3)$

#### y-intercept:
Set $x = 0$:
$$
y = -\frac{1}{3}(0 + 1)(0 - 5) = -\frac{1}{3}(1)(-5) = \frac{5}{3} \approx 1.67
$$
y-intercept: $(0, \frac{5}{3})$

#### One more point:
Try $x = 1$:
$$
y = -\frac{1}{3}(1 + 1)(1 - 5) = -\frac{1}{3}(2)(-4) = -\frac{1}{3}(-8) = \frac{8}{3} \approx 2.67 \Rightarrow (1, \frac{8}{3})$
$$

✔ Answer:
- x-intercepts: $(-1, 0), (5, 0)$
- Axis of symmetry: $x = 2$
- Vertex: $(2, 3)$
- y-intercept: $(0, \frac{5}{3})$
- Extra point: $(1, \frac{8}{3})$

---

5) $ y = 4(x + 2)(x + 1) $

#### x-intercepts:
Set $y = 0$:
$$
x = -2, x = -1
$$
x-intercepts: $(-2, 0), (-1, 0)$

#### Axis of symmetry:
$$
x = \frac{-2 + (-1)}{2} = \frac{-3}{2} = -1.5
$$

#### Vertex:
Plug $x = -1.5$:
$$
y = 4(-1.5 + 2)(-1.5 + 1) = 4(0.5)(-0.5) = 4(-0.25) = -1
$$
Vertex: $(-1.5, -1)$

#### y-intercept:
Set $x = 0$:
$$
y = 4(0 + 2)(0 + 1) = 4(2)(1) = 8
$$
y-intercept: $(0, 8)$

#### One more point:
Try $x = -3$:
$$
y = 4(-3 + 2)(-3 + 1) = 4(-1)(-2) = 4(2) = 8 \Rightarrow (-3, 8)
$$

✔ Answer:
- x-intercepts: $(-2, 0), (-1, 0)$
- Axis of symmetry: $x = -1.5$
- Vertex: $(-1.5, -1)$
- y-intercept: $(0, 8)$
- Extra point: $(-3, 8)$

---

6) $ y = -(x - 3)(x - 3) $

This is: $y = -(x - 3)^2$

#### x-intercepts:
Set $y = 0$:
$$
-(x - 3)^2 = 0 \Rightarrow x = 3
$$
Double root → x-intercept: $(3, 0)$

#### Axis of symmetry:
Since both roots are 3, axis is $x = 3$

#### Vertex:
Plug $x = 3$:
$$
y = -(3 - 3)^2 = 0 \Rightarrow (3, 0)
$$
Vertex: $(3, 0)$

Note: Vertex and x-intercept are the same because it touches the x-axis at the vertex.

#### y-intercept:
Set $x = 0$:
$$
y = -(0 - 3)^2 = -(9) = -9
$$
y-intercept: $(0, -9)$

#### One more point:
Try $x = 1$:
$$
y = -(1 - 3)^2 = -(-2)^2 = -4 \Rightarrow (1, -4)
$$

✔ Answer:
- x-intercepts: $(3, 0)$ (double root)
- Axis of symmetry: $x = 3$
- Vertex: $(3, 0)$
- y-intercept: $(0, -9)$
- Extra point: $(1, -4)$

---

✔ Final Answers Summary:

| Problem | x-intercepts | Axis of Symmetry | Vertex | y-intercept | Extra Point |
|--------|--------------|------------------|--------|-------------|-------------|
| 1 | $(-4, 0), (2, 0)$ | $x = -1$ | $(-1, -4.5)$ | $(0, -4)$ | $(1, -2.5)$ |
| 2 | $(0, 0), (8, 0)$ | $x = 4$ | $(4, 8)$ | $(0, 0)$ | $(2, 6)$ |
| 3 | $(-2, 0), (2, 0)$ | $x = 0$ | $(0, -4)$ | $(0, -4)$ | $(1, -3)$ |
| 4 | $(-1, 0), (5, 0)$ | $x = 2$ | $(2, 3)$ | $(0, \frac{5}{3})$ | $(1, \frac{8}{3})$ |
| 5 | $(-2, 0), (-1, 0)$ | $x = -1.5$ | $(-1.5, -1)$ | $(0, 8)$ | $(-3, 8)$ |
| 6 | $(3, 0)$ | $x = 3$ | $(3, 0)$ | $(0, -9)$ | $(1, -4)$ |

> 🔍 Note: For #2, I assumed the equation was $y = -\frac{1}{2}x(x - 8)$ due to the likely typo in the original. If the original was truly $y = -\frac{1}{2}(x - 8)$, then it's not quadratic — please verify the problem.

Let me know if you'd like me to graph any of these or provide visual explanations!