Question

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

Answer 1

The point on the curve is (–4, 140)

Solution:

Given $y=8x^2-3x$ and slope is –67.

slope = –67

$\frac{dy}{dx}=-67$  – – – – (1)

Now calculate $\frac{dy}{dx}$ for the given curve $y=8x^2-3x$.

Using differential rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

$\frac{dy}{dx}=\frac{d}{dx}(8x^2-3x)$

    $=\frac{d}{dx}(8x^2)-\frac{d}{dx}(3x)$

    $=(8\times2x^{2-1})-(3\times1x^{1-1})$

    $=16x^1-3x^0$

    $=16x-3$

$\frac{dy}{dx}=16x-3$ – – – – (2)

Equate (1) and (2).

$16x-3=-67$

$16x=-67+3$

$16x=-64$

⇒ x = –4

Substitute x = –4 in y.

⇒ $y=8(-4)^2-3(-4)$

⇒    $=8(16)+12$

⇒ y = 140

Hence, the point on the curve given is (–4, 140).