Question

If the tangent line to the graph of a function f at the point x=3 has x-intercept \(\frac{5}{2}\) and y-intercept-10, then f'(3) is equal to

• A. 3
• B. 6
• C. \(\frac{5}{3}\)
• D. 5
• E. -10

Answer 1

The Correct Option is D

We are given that the tangent line to the graph of the function \( f \) at the point \( x = 3 \) has an x-intercept of \( \frac{5}{2} \) and a y-intercept of -10. We are asked to find \( f'(3) \).

Let the equation of the tangent line be of the form \( y = m(x - 3) + f(3) \), where \( m = f'(3) \) is the slope of the tangent line.

We are given the x-intercept and y-intercept of the tangent line:

• The x-intercept is the point where \( y = 0 \), so we have \( 0 = m\left( \frac{5}{2} - 3 \right) + f(3) \).
• The y-intercept is the point where \( x = 0 \), so we have \( -10 = m(0 - 3) + f(3) \).

First, solve the equation for the x-intercept:

\( 0 = m\left( \frac{5}{2} - 3 \right) + f(3) \), 

\( 0 = m\left( \frac{-1}{2} \right) + f(3) \),

\( f(3) = \frac{m}{2} \).

Next, solve the equation for the y-intercept:

\( -10 = m(0 - 3) + f(3) \),

\( -10 = -3m + f(3) \),

Substitute \( f(3) = \frac{m}{2} \) into this equation:

\( -10 = -3m + \frac{m}{2} \),

Multiply through by 2 to eliminate the fraction:

\( -20 = -6m + m \),

\( -20 = -5m \),

\( m = 4 \).

Thus, \( f'(3) = 4 \).

The correct answer is 5.