Find the equation of the tangent line to the graph of the given function at the given value of x.
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Question

Quizplus · Accepted Answer

To find the equation of the tangent line, we need to find the derivative of the given function and evaluate it at the given value of x.

Taking the derivative of the function y = 2x^2 + 5x - 7, we get:

y' = 4x + 5

Evaluating this derivative at x = -3, we get:

y'(-3) = 4(-3) + 5 = -7

So the slope of the tangent line to the graph of the function at x = -3 is -7.

Now we need to find the y-intercept of the tangent line. We know that the point (-3, f(-3)) lies on both the function and the tangent line. So we can use this point to find the y-intercept of the tangent line.

Plugging in x = -3 into the function y = 2x^2 + 5x - 7, we get:

f(-3) = 2(-3)^2 + 5(-3) - 7 = -16

So the point of tangency is (-3, -16).

Using the point-slope form of the equation of a line, the equation of the tangent line is:

y - (-16) = -7(x - (-3))

Simplifying this equation, we get:

y = -7x + 5

Therefore, the equation of the tangent line to the graph of the given function at x = -3 is y = -7x + 5, which corresponds to choice C. Using the point-slope form of the equation of a line, we have:
y - f(x) = f'(x)(x - x)
Substituting x = 1, f(x) = -2, and f'(x) = 3, we get:
y - (-2) = 3(x - 1)
Simplifying, we get:
y = 3x - 5
Therefore, the equation of the tangent line is y = 3x - 5, which corresponds to choice C.

Find the Equation of the Tangent Line to the Graph