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Find the slope of the line tangent to the graph of (f) at x=2 of f(x)=3x^4−4x^2+7.
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Just substitute x=2 to the one you've differentiated
\[f'(x)=12x^3-8x \\ \\ f'(2)=12(2)^3-8(2) = 80\]
That's the slope/gradient,
do you how i can find the second part the tangent line 80 would be my slope but where do i get my y and x
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|dw:1365571474822:dw|
From the point slope formula, \[y-y_1=m(x-x_1)\] We need y1, we already known m and x1, m=slope using the value of "x", we can find "y" from the first equation let y=f(x) y=3x^4−4x^2+7 y=39 (2,39) m=80 \[y-y_1=m(x-x_1)\] \[y-39=80(x-2)\]
oh ok so i just get y by itself thanks
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