Find the standard (slope-intercept form) equation of the tangent line to the following functions at the specified points:

Question

Find the standard (slope-intercept form) equation of the tangent line to the following functions at the specified points:

a. 𝑓(𝑥) = 𝑥 3 − 2 at the point (2 , 6)

b. 𝑓(𝑥) = √25 − 𝑥^2 at the point where 𝑥 = 3

c. 𝑓(𝑥) = 𝑥^2 − √𝑥 at the point where 𝑥 = 1

Answer 1

y=x^3 - 2 at the point (2,6)

take the derivative to find the slope

y'= 3x^2 = slope

y'(2) = 3(2)^2 = 3(4) = 12 = slope when x=2

slope of the tangent line = 12

y=mx + b is the general equation for a line in slope intercept form, m=12

y=12x +b, plug in the point (2,6) to calculate b which is the y intercept

6=12(2)+b

b=6-24 = -18

y=12x -18 is the equation of the line tangent to x^3 -2 at the point (2,6)

f(x)=sqr(25-x^2)

y^2 =25-x^2

x^2+y^2 = 5^2 is a circle with radius 5 and center at the origin

but f(x) >0, so it's a semi-circle, the top half of the circle

at x=3, y= sqr(25-9) = +sqr16 = 4

sketch the graph of the semi-circle, the slope at (3,4) is downward sloping, m<0

y'(x) = (1/2)(25-x^2)^(-1/2)(-2x) = -x/(25-x^2)^(1/2)

y'(x)= -3/(25-9)^(1/2)= -3/sqr16 = -3/4

y=mx + b

y=(-3/4)x + b, plug in the point (3,4) to solve for b

4=(-3/4)(3) + b

b = 4+9/4 = 25/4

y=-3x/4 + 25/4 or

3x+4y = 25 is the equation of the tangent line to the semi-circle at the point (3,4)

f(x) = x^2 -sqrx x=1. y=1^2 -sqr1 = 1-1 =0 (x,y) = (1,0) is the point where the tangent line touches the curve

y'(x) = 2x -(1/2)x^(-1/2)

y'(1) = 2 - (1/2) = 3/2

y=mx +b

y=3x/2 +b, plug in the point (1,0) to solve for b

0=3/2 +b

b = -3/2

y-=3x/2 -3/2 or

3x+2y =-3 is the equation of the line tangent to x^2-sqrx at the point (1,0) where x=1