Let f(x) = 3 x^2 - 8. Find an equation of the line tangent to the graph of f at the point where x = -3. y =

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Let f(x) = 3 x^2 - 8. Find an equation of the line tangent to the graph of f at the point where x = -3. y =

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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y= -18?
y'=6x 6(-3)=y=-18
that's the slope, can you figure out the rest?

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plug -3 into the original equation, to find the corresponding y value, then use the (x,y) point you get with the slope m=-18 with the point slope formula y-y1=m(x-x1) to solve for the line
You have to find the point that corresponds to x = -3 on your graph f(x) We get: f(-3) = 3(-3)^2 -8 = 27 - 8 =19 Then you have to find the derivative of the function, to find the slope of the point at x = -3 Taking the derivative, f'(x) = 3*2x - 0 Then, f'(x) = 6x To find the slope at point x= -3, substitute x = -3 in. We get f'(-3) = -18 Now we now the point on f(x), which is (-3, 19), and the slope at x= -3, which is -18 Now we use the equation of the line formula which is y-y0 = m(x-x0), Now we just substitute our values in and get: y-19 = -18(x--3) y-19 = -18(x+3)

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