Question

introduction to derivatives: problem 3 (1 point) let f(5) = -9 and f’(5) = -6. then the equation of the tangent line to the graph of y = f(x) at x = 5 is y = \boxed{}

Explanation:

Step1: Recall the point - slope form of a line

The point - slope form of a line is given by \(y - y_1=m(x - x_1)\), where \((x_1,y_1)\) is a point on the line and \(m\) is the slope of the line.

For the tangent line to the graph of \(y = f(x)\) at \(x = 5\):

• The point \((x_1,y_1)\) on the tangent line (and also on the graph of \(y = f(x)\)) is \((5,f(5))\). We know that \(f(5)=- 9\), so \(x_1 = 5\) and \(y_1=-9\).
• The slope \(m\) of the tangent line to the graph of \(y = f(x)\) at \(x = 5\) is equal to the derivative of the function at that point, i.e., \(m = f^{\prime}(5)\). We know that \(f^{\prime}(5)=-6\).

Step2: Substitute the values into the point - slope form

Substitute \(x_1 = 5\), \(y_1=-9\) and \(m=-6\) into the point - slope form \(y - y_1=m(x - x_1)\):

\(y-(-9)=-6(x - 5)\)

Simplify the left - hand side: \(y + 9=-6(x - 5)\)

Then, distribute the \(-6\) on the right - hand side: \(y+9=-6x + 30\)

Subtract 9 from both sides to solve for \(y\): \(y=-6x+30 - 9\)

\(y=-6x + 21\)

Answer:

\(y=-6x + 21\)