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# The curves $y=4x^2+2x-8$ and $y=x^3-x+13$ touch each other at the point_________.

Toolbox:
• Slope of a line is $ax+by+c=0$ is $- \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
• If $y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p =$ slope of the normal to $y = f(x)$ at point $p$
Step 1
Let curve $c_1$ be $y=4x^2+2x-8$
Let curve $c_2$ be $y=x^3-x+13$
Differentiating curve $c_1$ w.r.t $x$ we get,
$\large\frac{dy}{dx}=8x+2$
Differentiating curve $c_2$ w.r.t $x$ we get,
$\large\frac{dy}{dx}=3x^2-1$
Since they touch each other, let us assume that the slope of the tangent are equal at $(x_1, y_1)$
i.e., $8x_1+2 = 3x_1^2-1$
$\Rightarrow 3x_1^2-8x_1-3=0$
On factorizing we get,
$(3x_1+1)(x-3), \: \Rightarrow x_1=-\large\frac{1}{3} \: and \: x=3$
Since $x_1 \neq - \large\frac{1}{3}$, let us consider $x = 3$
$\therefore y_1 = 4(3)^2+2(3)-8$
$= 34$
Hence the point of intersection is $(3, 34)$