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. $y= 4x^(2) +1$

Answers

Answer 1

Answer:

Final answer:

To find the equation of a line tangent to the given curve, we need to find the derivative of the curve at the point of tangency and substitute the x-coordinate of the point of tangency to find the slope. The equation of the tangent line with 0 slope and y-intercept of 8 is y = 8.

Explanation:

To find the equation of a line tangent to a curve, we need to find the derivative of the curve at the point of tangency.

The given curve is y = 4x² + 1 (equation 1).

First, find the derivative of equation 1, which gives us dy/dx = 8x (equation 2).

Next, substitute the x-coordinate of the point of tangency into equation 2 to find the slope of the tangent line.

Since the line cuts the y-axis at (0,8), the x-coordinate of the point of tangency is 0.

Substituting x=0 into equation 2, we get the slope of the tangent line as m = 8(0) = 0.

The equation of a line in the form y = mx + c, where m is the slope and c is the y-intercept.

Since the slope of the tangent line is 0, the equation of the tangent line is y = 0x + c. And since the line cuts the y-axis at (0,8), the y-intercept is 8.

Therefore, the equation of the tangent line is y = 8.

Learn more about Equation of a tangent line here:

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Answer 2

Answer:

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:

$f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -(7)/(4)$

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

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Answer:

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step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

Given expression is $(12+r)^3$

To find how many terms in the expansion of the given expression :

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Answers

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Answers

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