Question

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

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Answer 1

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):

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