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Dvinal [7]
3 years ago
11

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

Mathematics
1 answer:
AleksandrR [38]3 years ago
3 0

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).

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