Question

introduction to derivatives: problem 3 (1 point) let f(-7) = -1 and f(-7) = 7, then the equation of the tangent line to the graph of y = f(x) at x = -7 is y = you have attempted this problem 0 times. you have unlimited attempts remaining.

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 \(x=a\), the point \((x_1,y_1)\) is \((a,f(a))\) and the slope \(m\) of the tangent line is \(f^{\prime}(a)\).

Here, \(a=-7\), \(f(-7)=-1\) (so \(x_1=-7\) and \(y_1 = - 1\)) and \(f^{\prime}(-7)=7\) (so the slope \(m = 7\)).

Step2: Substitute the values into the point - slope form

Substitute \(x_1=-7\), \(y_1=-1\) and \(m = 7\) into the point - slope form \(y - y_1=m(x - x_1)\):

\(y-(-1)=7(x - (-7))\)

Simplify the left - hand side and the right - hand side:

\(y + 1=7(x + 7)\)

Step3: Expand and simplify the equation

Expand the right - hand side: \(y+1 = 7x+49\)

Subtract 1 from both sides to get the equation in slope - intercept form (\(y=mx + b\)):

\(y=7x+49 - 1\)

\(y=7x + 48\)

Answer:

\(y = 7x+48\)