Step-by-step explanation:
step explanation:
We need to find the equation of the tangent lines of Points A and B.
Differentiate the equation:
\displaystyle \frac{dy}{dx}=3x^2-3
dx
dy
=3x
2
−3
Point A has an x-coordinate of -1. Hence, the slope of its tangent line is:
\displaystyle \frac{dy}{dx}\Big|_{x=-1}=3(-1)^2-3=0
dx
dy
∣
∣
∣
∣
x=−1
=3(−1)
2
−3=0
Find the y-coordinate of Point A using the original equation:
y(-1)=(-1)^3-3(-1)=2y(−1)=(−1)
3
−3(−1)=2
Hence, Point A is (-1, 2).
Thus, the tangent line at Point A is:
y-2=0(x-(-1))y−2=0(x−(−1))
Simplify:
y=2y=2
Point B has an x-coordinate of 4. Hence, the slope of its tangent line is:
\displaystyle \frac{dy}{dx}\Big|_{x=4}=3(4)^2-3=45
dx
dy
∣
∣
∣
∣
x=4
=3(4)
2
−3=45
Find the y-coordinate of Point B:
y(4)=(4)^3-3(4)=52y(4)=(4)
3
−3(4)=52
Thus, Point B is at (4, 52).
So, the tangent line at Point B is:
y-52=45(x-4)y−52=45(x−4)
Simplify:
y=45x-128y=45x−128
Point C occurs at the intersections of the tangent lines of Points A and B. Set the two equations equal to each other and solve for x:
2=45x-1282=45x−128
We acquire:
\displaystyle x=\frac{26}{9}x=
9
26
Since one of our equations is y = 2, the y-coordinate is 2.
Hence, Point C is:
\displaystyle C=\left(\frac{26}{9}, 2\right)C=(
9
26
,2)