What is the equation of the line tangent to the function f(x) = 4x^2 + 5x at the point (-2, 6). Will give brainliest! Thank you.

Question

What is the equation of the line tangent to the function f(x) = 4x^2 + 5x at the point (-2, 6).

Will give brainliest! Thank you.

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Hưng Khoa 3 years 2021-08-31T13:13:55+00:00 1 Answers 3 views 0

Answers ( )

  1. phucdien
    0
    2021-08-31T13:15:08+00:00
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    Answer:

    y=-11x-16

    Step-by-step explanation:

    We want to find the equation of the tangent line to the function:

    f(x)=4x^2+5x

    At the point (-2, 6).

    First, we will need the slope of the tangent line. So, differentiate* the function:

    f'(x)=8x+5

    Find the slope when x = -2:

    f'(-2)=8(-2)+5=-11

    Now, we can use the point-slope form:

    y-y_1=m(x-x_1)

    Our point is (-2, 6) and our slope is -11. Substitute:

    y-(6)=-11(x-(-2))

    Simplify:

    y-6=-11(x+2)

    Distribute:

    y-6=-11x-22

    And add six to both sides. Therefore, our equation is:

    y=-11x-16

    If you have not yet learned differentiation, here’s the method using the difference quotient! The difference quotient is given by:

    \displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

    Here, x = -2. Substitute:

    \displaystyle f'(-2)=\lim_{h\to 0}\frac{f(-2+h)-f(-2)}{h}

    Substitute (we are given the point (-2, 6). So, f(-2) = 6).

    \displaystyle f'(-2)=\lim_{h\to 0}\frac{(4(-2+h)^2+5(-2+h))-(6)}{h}

    Expand and simplify:

    \displaystyle f'(-2)=\lim_{h\to 0}\frac{(4(4-4h+h^2)+(-10+5h))-(6)}{h}

    Distribute:

    \displaystyle f'(-2)=\lim_{h\to 0}\frac{16-16h+4h^2-10+5h-6}{h}

    Simplify:

    \displaystyle f'(-2)=\lim_{h\to 0}\frac{4h^2-11h}{h}

    Evaluate the limit (using direct substitution):

    \displaystyle f'(-2) = \lim_{h\to 0}4h-11=4(0)-11=-11

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