Question

for what point on the curve of y=8x^2 - 3x is the slope of a tangent line equal to -67

Answer 1

Answer:

(-4,140)

Step-by-step explanation:

We are given;

Equation of a curve; $y=8x^2-3x$

Slope of a tangent to the curve = -67

We are required to determine the point on the curve whose tangent has a slope of -67

We are going to first get the gradient function for the curve;

$y=8x^2-3x\\\\ \frac{dy}{dx}=16x-3$

We can use the gradient function to determine the value of x

$\frac{dy}{dx}=16x-3\\Then;\\-67= 16x-3\\-64=16x\\x=-4$

Thus, the value of X at the point is 4

We can use the equation of the curve to determine the value of y

$y=8x^2-3x\\y=8(-4)^2-3(-4)\\ = 140$

Therefore, the point will be (-4,140)