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I am having trouble with the following question: Given, $y=x^{3}-2x+3$, Find the equation of the tangent at at x =2.

I tried doing $y_{2}-y_{1}=m(x_{2}-x_{1})$. Then I plugged 2 into the equation and got 7 which I used as my m value but I am not really sure what to do.

Kasper Verhulst
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Timmy
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3 Answers3

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The tangent at $f(x)$ in $x=x_0$ is given by

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

in this case $f(2)=7$ and $f'(2)=10$ thus

$$y=7+10(x-2)\implies y=10x-13$$

user
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The slope of the tangent is given by the derivative, so you need to take $\frac {dy}{dx}$ and evaluate it at $x=2$ to get the slope. Then $(x_1,y_1)$ is the point on the curve at $x=2$ and you have your linear equation.

Ross Millikan
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  • I have never heard of derivatives before. I am in grade 12, is derivitives supposed to be taught in grade 12? – Timmy Jan 18 '18 at 22:59
  • If you are asked this question they should have been taught. It is a big enough subject you wouldn't have missed it. – Ross Millikan Jan 18 '18 at 23:01
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We know that $y=x^3-2x+3$.

Take the derivative: $$\frac{dy}{dx}=3x^2-2$$

Substitute $x=2$: $$\frac{dy}{dx}=3\cdot2^2-2=10$$

So, the slope is $10$. Consider that it passes the curve at $x=2\rightarrow y=7$. Therefore, the tangent is $(2,7)$ and the equation is $$y-y_1=m(x-x_1)\\ y-7=10(x-2)\\ y=10x-20+7\\ y=10x-13$$

user_194421
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