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anonymous

  • 4 years ago

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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  1. anonymous
    • 4 years ago
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    y= -18?

  2. anonymous
    • 4 years ago
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    y'=6x 6(-3)=y=-18

  3. anonymous
    • 4 years ago
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    that's the slope, can you figure out the rest?

  4. anonymous
    • 4 years ago
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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

  5. anonymous
    • 4 years ago
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    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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