Question

a) graph the function ( f(x) = x^2 + 5x )
b) draw the tangent lines to the graph at the points whose x-coordinates are -4, -3, and -1
c) find ( f(x) ) by determining ( lim_{h \to 0} \frac{f(x + h) - f(x)}{h} )
d) find ( f(-4) ), ( f(-3) ), and ( f(-1) ). these slopes should match those of the lines in part (b)
b) choose the correct tangent lines to the graph
c) ( f(x) = square )

Explanation:

Part c)

Step 1: Substitute \( f(x+h) \) and \( f(x) \)

Given \( f(x) = x^2 + 5x \), then \( f(x+h)=(x+h)^2 + 5(x+h) \). Expand: \( (x^2 + 2xh + h^2)+5x + 5h \).

Step 2: Compute \( f(x+h)-f(x) \)

\( (x^2 + 2xh + h^2 + 5x + 5h)-(x^2 + 5x)=2xh + h^2 + 5h \).

Step 3: Divide by \( h \)

\( \frac{f(x+h)-f(x)}{h}=\frac{2xh + h^2 + 5h}{h}=2x + h + 5 \) (for \( h
eq0 \)).

Step 4: Take the limit as \( h\to0 \)

\( \lim_{h\to0}(2x + h + 5)=2x + 5 \).

Step 1: Find \( f'(-4) \)

Substitute \( x = -4 \) into \( f'(x)=2x + 5 \): \( 2(-4)+5=-8 + 5=-3 \).

Step 2: Find \( f'(-3) \)

Substitute \( x = -3 \): \( 2(-3)+5=-6 + 5=-1 \).

Step 3: Find \( f'(-1) \)

Substitute \( x = -1 \): \( 2(-1)+5=-2 + 5=3 \).

Answer:

\( f'(x)=2x + 5 \)

Part d)