Question

refer to the graph of ( y = f(x) = x^2 + x ) shown.
a) find the slope of the secant line joining ( (-1, f(-1)) ) and ( (3, f(3)) ).
b) find the slope of the secant line joining ( (-1, f(-1)) ) and ( (-1 + h, f(-1 + h)) ).
c) find the slope of the graph at ( (-1, f(-1)) ).
d) find the equation of the tangent line to the graph at ( (-1, f(-1)) ).

Explanation:

Part (a)

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

First, substitute \( x = -1 \) and \( x = 3 \) into \( f(x)=x^{2}+x \).
For \( x=-1 \): \( f(-1)=(-1)^{2}+(-1)=1 - 1 = 0 \)
For \( x = 3 \): \( f(3)=3^{2}+3=9 + 3 = 12 \)

Step 2: Use the slope formula for secant line

The slope \( m \) of the line joining two points \( (x_1,y_1) \) and \( (x_2,y_2) \) is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Here, \( (x_1,y_1)=(-1,f(-1))=(-1,0) \) and \( (x_2,y_2)=(3,f(3))=(3,12) \).
So, \( m=\frac{12-0}{3 - (-1)}=\frac{12}{4}=3 \)

Step 1: Find \( f(-1 + h) \)

Substitute \( x=-1 + h \) into \( f(x)=x^{2}+x \):
\( f(-1 + h)=(-1 + h)^{2}+(-1 + h) \)
Expand \( (-1 + h)^{2}=1-2h+h^{2} \), so \( f(-1 + h)=1-2h+h^{2}-1 + h=h^{2}-h \)

Step 2: Use the slope formula

The two points are \( (-1,f(-1))=(-1,0) \) and \( (-1 + h,f(-1 + h))=(-1 + h,h^{2}-h) \)
Slope \( m=\frac{(h^{2}-h)-0}{(-1 + h)-(-1)}=\frac{h^{2}-h}{h}=\frac{h(h - 1)}{h}=h - 1 \) (for \( h
eq0 \))

Step 1: Recall the definition of the slope of the graph at a point

The slope of the graph of \( y = f(x) \) at \( x = a \) is the limit of the slope of the secant line as \( h
ightarrow0 \), i.e., \( \lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h} \)
Here, \( a=-1 \), and from part (b), the slope of the secant line joining \( (-1,f(-1)) \) and \( (-1 + h,f(-1 + h)) \) is \( h - 1 \)

Step 2: Find the limit as \( h

ightarrow0 \)
\( \lim_{h
ightarrow0}(h - 1)=0 - 1=-1 \)

Answer:

The slope of the secant line is \( 3 \)

Part (b)