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kupik [55]
3 years ago
15

The line L is a tangent to the curve with equation y= 4x^2 +1 . The line L cuts the y axis at (0,8) and has a positive gradient.

Find the equation of L in the form y = mx + c.y= 4x^{2} +1
Mathematics
1 answer:
sergeinik [125]3 years ago
5 0

A generic point on the graph of the curve has coordinates

(x, 4x^2+1)

The derivative gives us the slope of the tangent line at a given point:

f(x) = 4x^2+1 \implies f'(x) = 8x

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through (k, 4k^2+1) and have slope 8k

So, we can write its equation using the point-slope formula: a line with slope m passing through (x_0, y_0) has equation

y-y_0 = m(x-x_0)

In this case, (x_0, y_0)=(k, 4k^2+1) and m=8k, so the equation becomes

y-4k^2-1 = 8k(x-k)

We can rewrite the equation as follows:

y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1

We know that this function must give 0 when evaluated at x=0:

f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}

This equation has no real solution, so the problem looks impossible.

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