Question

which function has zeros of - 4 and 2?
a. $f(x) = x^2 + 7x - 8$
b. $g(x) = x^2 - 7x - 8$
c. image of a parabola
d. image of a parabola

Explanation:

Step1: Recall the factor theorem

If a function has zeros at \( x = a \) and \( x = b \), then the function can be written as \( f(x)=(x - a)(x - b) \). Here, the zeros are \( - 4 \) and \( 2 \), so \( a=-4 \), \( b = 2 \). Then the function is \( f(x)=(x+4)(x - 2) \).

Step2: Expand the function

Expand \( (x + 4)(x - 2) \):

$$
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$$

Wait, maybe we can also check the zeros of the given functions by substituting \( x=-4 \) and \( x = 2 \) into each function.

Step3: Check option a

For \( f(x)=x^{2}+7x - 8 \), when \( x=-4 \):
\( f(-4)=(-4)^{2}+7\times(-4)-8=16-28 - 8=-20
eq0 \)
When \( x = 2 \):
\( f(2)=2^{2}+7\times2-8=4 + 14-8 = 10
eq0 \)

Step4: Check option b

For \( g(x)=x^{2}-7x - 8 \), when \( x=-4 \):
\( g(-4)=(-4)^{2}-7\times(-4)-8=16 + 28-8 = 36
eq0 \)
When \( x = 2 \):
\( g(2)=2^{2}-7\times2-8=4-14 - 8=-18
eq0 \)

Step5: Check the graphs

For a quadratic function \( y = ax^{2}+bx + c \) with zeros at \( x=-4 \) and \( x = 2 \), the graph should intersect the x - axis at \( x=-4 \) (left side of the y - axis) and \( x = 2 \) (right side of the y - axis).

• Graph c: The x - intercepts seem to be symmetric about the y - axis? No, wait, looking at the grid, the left x - intercept is at \( x=-2 \) maybe? No, let's check graph d.
• Graph d: The left x - intercept is at \( x=-4 \) (since from the y - axis, moving 4 units left) and the right x - intercept is at \( x = 2 \) (moving 2 units right). So the graph of option d has zeros at \( x=-4 \) and \( x = 2 \).

Answer:

d. The graph in option d (the lower graph)