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Question

Find the equation of the tangent to the curve $y=(1+2x)^2$ at the point (1,9)

I know that I have to derive the function that is given but then that spits out the slope only. I dont know how to generate an equation that has the same slope while having a point (1,9)

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    A tangent through $(1,9)$ has the equation $y=m(x-1)+9$ and you know $m$ through the derivative at $x=1$. – 2017-01-22
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    hint: $slope = \Delta y / \Delta x$ – 2017-01-23

2 Answers 2

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Use the point slope formula.

This says that the line passing through $(x_1,y_1)$ with slope $m$ can be written as $y - y_1= m(x-x_1)$.

The derivative of $(1+2x)^2$ is $4(1+2x)$, and evaluating this for $x=1$ gives us $m=12$.

Therefore, the tangent line is $y-9=12(x-1)$, or $y=12x-3$.

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You can write,

$$y=mx+b$$

Then substitute $m$, $x=1$ and $y=9$ and solve for $b$. Or more efficiently notice that,

$$y=m(x-1)$$

Is a line with slope $m$, but when we plug $x=1$ in we get $y=0$ but we want $y=9=0+9$. So we have to add $9$ to this equation to get what we want.

$$y=m(x-1)+9$$