Question

find the indicated quantities for f(x) = 3x². (a) the slope of the secant line through the points (1, f(1)) and (1 + h, f(1 + h)), h ≠ 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 ≠ 0, is 6 + 3h. (b) the slope of the graph at (1, f(1)) is 6. (type an integer or a simplified fraction.) (c) the equation of the tangent line at (1, f(1)) is y = ☐.

Explanation:

Step1: Find f(1)

First, calculate \( f(1) \) for the function \( f(x) = 3x^2 \). Substitute \( x = 1 \) into the function:
\( f(1) = 3(1)^2 = 3 \). So the point is \( (1, 3) \).

Step2: Use point - slope form

The slope of the tangent line at \( (1, f(1)) \) is 6 (from part B). The point - slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1)=(1,3) \) and \( m = 6 \).

Substitute these values into the point - slope formula:
\( y - 3=6(x - 1) \)

Step3: Simplify the equation

Expand the right - hand side: \( y - 3=6x-6 \)

Then, add 3 to both sides of the equation to solve for \( y \):
\( y=6x - 6 + 3=6x-3 \)

Answer:

\( 6x - 3 \)