Question

consider the function ( f(x) = -x^2 - 3x + 8 ).
(a) what is the equation of the secant line that intersects the graph of ( f(x) ) when ( x = -4 ) and ( x = 0 )? (enter an equation in terms of ( x ) and ( y ).)

(b) what is the equation of the secant line that intersects the graph of ( f(x) ) when ( x = -1 ) and ( x = 1 )? (enter an equation in terms of ( x ) and ( y ).)

(c) which of the following is a graph of the function ( f(x) ), including the two secant lines found in parts (a) and (b)?

Explanation:

Step1: Find f(-4) and f(0)

$f(-4) = -(-4)^2 -3(-4)+8 = -16+12+8=4$
$f(0) = -(0)^2 -3(0)+8=8$

Step2: Calculate slope for (a)

$m_a = \frac{f(0)-f(-4)}{0-(-4)} = \frac{8-4}{4} = 1$

Step3: Find secant line (a)

Use point-slope form $y-y_1=m(x-x_1)$ with $(0,8)$:
$y-8=1(x-0) \implies y=x+8$

Step4: Find f(-1) and f(1)

$f(-1) = -(-1)^2 -3(-1)+8 = -1+3+8=10$
$f(1) = -(1)^2 -3(1)+8 = -1-3+8=4$

Step5: Calculate slope for (b)

$m_b = \frac{f(1)-f(-1)}{1-(-1)} = \frac{4-10}{2} = -3$

Step6: Find secant line (b)

Use point-slope form $y-y_1=m(x-x_1)$ with $(1,4)$:
$y-4=-3(x-1) \implies y=-3x+7$

Step7: Match graph to functions

$f(x)=-x^2-3x+8$ is a downward-opening parabola. Line $y=x+8$ has positive slope, line $y=-3x+7$ has negative slope; both intersect the parabola at the correct points, matching the first (left) graph.

Answer:

(a) $y = x + 8$
(b) $y = -3x + 7$
(c) The first (left-hand) WebAssign Plot