Question

Mr. Walker gave his class the function f(x) = (x + 3)(x + 5). Four students made a claim about the function. Each student’s claim is below.

Jeremiah: The y-intercept is at (15, 0).
Lindsay: The x-intercepts are at (–3, 0) and (5, 0).
Stephen: The vertex is at (–4, –1).
Alexis: The midpoint between the x-intercepts is at (4, 0).


Which student’s claim about the function is correct?

The claim by
is correct.

Answer 1

Let's evaluate each student's claim regarding the function \( f(x) = (x + 3)(x + 5) \).

1. **Jeremiah:** The y-intercept is at (15, 0).
- To find the y-intercept, we set \( x = 0 \):
\[
f(0) = (0 + 3)(0 + 5) = 3 \times 5 = 15.
\]
- The y-intercept is at (0, 15), not (15, 0). **Jeremiah's claim is incorrect.**

2. **Lindsay:** The x-intercepts are at (–3, 0) and (5, 0).
- To find the x-intercepts, we set \( f(x) = 0 \):
\[
(x + 3)(x + 5) = 0.
\]
- This gives us the solutions:
\[
x + 3 = 0 \implies x = -3,
\]
\[
x + 5 = 0 \implies x = -5.
\]
- Therefore, the x-intercepts are (–3, 0) and (–5, 0). **Lindsay's claim is incorrect.**

3. **Stephen:** The vertex is at (–4, –1).
- To find the vertex of a quadratic function in the form \( ax^2 + bx + c \), we can convert \( f(x) \) to standard form or use the vertex formula \( x = -\frac{b}{2a} \).
- First, let's expand \( f(x) \):
\[
f(x) = x^2 + 8x + 15.
\]
- Here, \( a = 1 \), \( b = 8 \), and \( c = 15 \). The x-coordinate of the vertex is:
\[
x = -\frac{8}{2 \times 1} = -4.
\]
- Finding the y-coordinate of the vertex:
\[
f(-4) = (-4 + 3)(-4 + 5) = (-1)(1) = -1.
\]
- Therefore, the vertex is at (–4, –1). **Stephen's claim is correct.**

4. **Alexis:** The midpoint between the x-intercepts is at (4, 0).
- The x-intercepts are \( x = -3 \) and \( x = -5 \).
- The midpoint \( m \) of two points \( x_1 \) and \( x_2 \) is given by:
\[
m = \frac{x_1 + x_2}{2} = \frac{-3 + (-5)}{2} = \frac{-8}{2} = -4.
\]
- Thus, the midpoint of the x-intercepts is at (–4, 0), not (4, 0). **Alexis's claim is incorrect.**

Therefore, the claim by **Stephen** is correct.