Answer to Question #291463 in Calculus for nickname

Question

For each problem, find the equation of the line tangent to the function at the given point. Your answer should be a slope-form.

1) y = 2x^2 + 16x + 32 at (-3, 2)

2) y = (x-2)^(1/3) at (3,1)

3) y = -x^3 + 3x^2 -4 at (3, -4)

4) y = -3/ x^2 - 4 at (-1, 1)

Answer 1

1)

$$y'=4x+16$$

$$slope=m=y'(-3)=4(-3)+16=4$$

$$y-2=4(x-(-3))$$

The equation of the line tangent to the function at $(-3, 2)$ is

$$y=4x+14$$

2)

$$y'=\dfrac{1}{3(x-2)^{2/3}}$$

$$slope=m=y'(3)=\dfrac{1}{3(3-2)^{2/3}}=\dfrac{1}{3}$$

$$y-1=\dfrac{1}{3}(x-3)$$

The equation of the line tangent to the function at $(3, 1)$ is

$$y=\dfrac{1}{3}x$$

3)

$$y'=-3x^2+6x$$

$$slope=m=y'(3)=-3(3)^2+6(3)=-9$$

$$y-(-4)=-9(x-3)$$

The equation of the line tangent to the function at $(3, -4)$ is

$$y=-9x+23$$

4)

$$y'=\dfrac{6x}{(x^2-4)^{2}}$$

$$slope=m=y'(-1)=\dfrac{6(-1)}{((-1)^2-4)^{2}}=-\dfrac{2}{3}$$

$$y-1=-\dfrac{2}{3}(x-(-1))$$

The equation of the line tangent to the function at $(-1, 1)$ is

$$y=-\dfrac{2}{3}x+\dfrac{1}{3}$$