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I have to find the tangent lines to the function $$f(x)=(x+1)^{3/2}$$ that pass through the point $(4/3, 3)$. I found the derivative of the function $$f'(x)=\frac{3\sqrt{x+1}}{2}$$ and know that that will be the slope of the two lines. I think I use the point-slope form to get the equation $f(x)-3=f'(x)(x-4/3)$. From here I try solving for $x$ and get absolutely nowhere.

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At the point $\left(t,(t+1)^{\frac 32}\right), \ f'(t)= \frac 32 \sqrt{t+1}$. The equation of the tangent to the curve at that point is $y = \frac 32 \sqrt{t+1}(x - t) + (t+1)^{\frac 32}$.

Now let $x=\frac 43$and $y=3$. Then solve for $t$. You should get two nice integers for answers.

\begin{align} \frac 32 \sqrt{t+1}\left(\frac 43 - t \right) + (t+1)^{\frac 32} &= 3\\ \sqrt{t+1}(4-3t) + 2(t+1)^{\frac 32} &= 6 \\ (4-3t) + 2(t+1) &= \frac{6}{\sqrt{t+1}} \\ 6-t &= \frac{6}{\sqrt{t+1}} \\ etc \end{align}

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  • $\begingroup$ Thanks bud. Big help. $\endgroup$ – user17763 Oct 12 '17 at 16:48
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A line passing through $\left(\frac 43,3\right)$ with slope $m$ has the following equation:$$y-3=m\left(x-\frac 43\right)$$

Now, if such a line passes through point $(x_1, y_1)$ on curve $f(x)=(x+1)^\frac 32$, it satisfies $$y_1^2=(x_1+1)^3,$$ and $$f'(x_1)=m=\frac{3\sqrt{x_1+1}}{2}.$$

You have unknowns $x_1,y_1, and\ m$. You also have three equations staisfied by the same. Can you now solve the problem?

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