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For the function $f(x)=x^2$ and $f'(x)=2x$ find the equation of the tangent line(s) to $f(x)$ from $(\frac{3}{4}, -1)$

I just started basic derivatives in my calc class. Can someone explain this to me? I set up the value a to represent the point on the parabola $y=x^2$ where the tangent line passes through making the point $(a, a^2)$. I then used point slope form to try and find the $x$ value on the parabola $y=2x(x-8) + a^2$ then I plugged in $\frac{3}{4}$ in $a^2 -\frac{3}{2}a + \frac{9}{8}$ but I struggling to find a coherent answer from this

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    $\begingroup$ Your procedure started out right. Let the point of tangency be $(a,a^2)$. The equation of the tangent line turns out to be $y-a^2=2a(x-a)$, or equivalently $y=2ax-a^2$. So $-1=2(3/4)a-a^2$. Multiply through by $2$ to get rid of fractions, and simplify a bit. We get $2a^2-3a-2=0$. Solve, using the Quadratic Formula, though we can also factor as $(2a+1)(a-2)$. $\endgroup$ – André Nicolas Oct 1 '15 at 0:52
  • $\begingroup$ Thank you for explaining this out for me! $\endgroup$ – Rach12 Oct 1 '15 at 0:53
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    $\begingroup$ You are welcome. Note that the slope is $2a$. I am not sure what you mean by point-slope form, possibly $\frac{y-a^2}{x-a}=2a$, which simplifies to what I wrote. $\endgroup$ – André Nicolas Oct 1 '15 at 0:54
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    $\begingroup$ Before doing any calculation, either draw the parabola or at least clearly visualize it. Draw the point $P=(3/4,-1)$, It is not on the parabola. You can see that there are two tangents to the parabola through $P$. You can even roughly estimate their slopes, to check against the result you will obtain later. $\endgroup$ – André Nicolas Oct 1 '15 at 1:31
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$$y - f(x_0)=f'(x_0)(x-x_0)$$

$$y-(-1)=[2(\frac{3}{4})](x-\frac{3}{4})$$

$$y+1=\frac{3}{2}x-\frac{9}{8}$$

$$y=\frac{3}{2}x-\frac{17}{8}$$

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$$y - f(x_0)=f'(x_0)(x-x_0)$$

That's about it.

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  • $\begingroup$ Thank you, but I am struggling to obtain the right values for "a" from point slope form $\endgroup$ – Rach12 Oct 1 '15 at 0:51

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