Lesson Explainer: Equations Of Tangent Lines And Normal ... - Nagwa

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Example 1: Finding the Equation of the Tangent to the Curve of a Polynomial Function at a Given Value for 𝑥

Find the equation of the tangent to the curve 𝑦=−2𝑥+8𝑥−19 at 𝑥=2.

Answer

To find the equation of the tangent to a curve at a point, we need two pieces of information: the coordinates of the point and the slope of the curve at this point.

The question wants us to find the tangent when 𝑥=2, so the 𝑥-coordinate is 2. We can find the 𝑦-coordinate for this value of 𝑥 by substituting 𝑥=2 into the equation of our curve: 𝑦=−2(2)+8(2)−19=−16+32−19=−3.

This gives us the point (2,−3) on our curve that our tangent line must pass through.

Next, we need the slope of the curve when 𝑥=2; to find this we need to differentiate: dddd𝑦𝑥=𝑥−2𝑥+8𝑥−19=−6𝑥+16𝑥.

We then substitute in 𝑥=2 to find the slope of the tangent line at this point: dd𝑦𝑥|||=−6(2)+16(2)=−24+32=8.

Now, to find the equation of our tangent line, we recall that the equation of a line of slope 𝑚 passing through (𝑥,𝑦) is given by 𝑦−𝑦=𝑚(𝑥−𝑥).

In our case, we have 𝑦−(−3)=8(𝑥−2)𝑦+3=8𝑥−16𝑦−8𝑥+19=0.

Therefore, the equation of the tangent line to our curve at 𝑥=2 is given by the equation 𝑦−8𝑥+19=0.

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