# How do you show the line #y=6x-5# is a tangent to the quadratic equation #f(x)= 5x^2 - 14x + 15#?

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To show that the line y=6x-5 is a tangent to the quadratic equation f(x)= 5x^2 - 14x + 15, we need to find the point of tangency.

First, we find the derivative of the quadratic equation f(x) to get f'(x).

f'(x) = 10x - 14

Next, we set f'(x) equal to the slope of the line y=6x-5, which is 6.

10x - 14 = 6

Solving this equation, we find x = 2.

Substituting x=2 into the quadratic equation f(x), we find f(2) = 5(2)^2 - 14(2) + 15 = 5.

Therefore, the point of tangency is (2, 5).

To confirm that the line y=6x-5 is a tangent, we can check if the slope of the tangent line at (2, 5) is indeed 6.

Taking the derivative of f(x) and evaluating it at x=2, we get f'(2) = 10(2) - 14 = 6.

Since the slope of the tangent line matches the slope of the line y=6x-5, we can conclude that the line y=6x-5 is a tangent to the quadratic equation f(x)= 5x^2 - 14x + 15 at the point (2, 5).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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