Question

find the indicated quantities for ( f(x) = 3x^2 ). (a) the slope of the secant line through the points ( (1,f(1)) ) and ( (1 + h,f(1 + h)) ), ( h
eq 0 ) (b) the slope of the graph at ( (1,f(1)) ) (c) the equation of the tangent line at ( (1,f(1)) ) (a) the slope of the secant line through the points ( (1,f(1)) ) and ( (1 + h,f(1 + h)) ), ( h
eq 0 ), is (square).

Explanation:

Step 1: Recall the slope formula

The slope \( m \) of the line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Here, \( x_1 = 1,y_1=f(1)\), \( x_2=1 + h,y_2=f(1 + h) \). So we first need to find \( f(1) \) and \( f(1 + h) \) for the function \( f(x)=3x^{2} \).

Step 2: Calculate \( f(1) \) and \( f(1 + h) \)

For \( f(1) \), substitute \( x = 1 \) into \( f(x) \):
\( f(1)=3\times(1)^{2}=3 \)
For \( f(1 + h) \), substitute \( x=1 + h \) into \( f(x) \):
\( f(1 + h)=3\times(1 + h)^{2}=3\times(1 + 2h+h^{2})=3 + 6h+3h^{2} \)

Step 3: Calculate the slope of the secant line

Using the slope formula \( m=\frac{f(1 + h)-f(1)}{(1 + h)-1} \)
Substitute \( f(1 + h)=3 + 6h + 3h^{2} \) and \( f(1) = 3 \) into the formula:
\( m=\frac{(3 + 6h+3h^{2})-3}{h}=\frac{6h + 3h^{2}}{h} \)
Since \( h
eq0 \), we can factor out \( h \) from the numerator and cancel with the denominator:
\( \frac{h(6 + 3h)}{h}=6 + 3h \)

Answer:

\( 6 + 3h \)