Kienastsrx

2022-03-16

What is the equation of the tangent line of $f\left(x\right)={x}^{2}-2x$ at x=5?

Jarrett Chan

Beginner2022-03-17Added 6 answers

So at that point on the curve, as it is a tangent the gradient will be the same.

The gradient is calculated by differentiating.

$\frac{dy}{dx}{x}^{2}-2xdx=\frac{1}{2}x-2$

We can then substitute x in to find the gradient at x=5.

Therefore,$\frac{1}{2}\left(5\right)-2=\frac{1}{2}$ .

This is our gradient.

Then we need to find y at the point x=5.

Substitute into${x}^{2}-2x$ to get ${\left(5\right)}^{2}-2\left(5\right)=15$ .

Now we can use the equation of$y-{y}_{1}=m(x-{x}_{1})$ .

Therefore:

$y-15=\frac{1}{2}(x-5)$

Expand and rearrange:

$y-15=\frac{1}{2}x-\frac{5}{2}$

$y=\frac{1}{2}x+\frac{25}{2}$

2y=x+25.

The gradient is calculated by differentiating.

We can then substitute x in to find the gradient at x=5.

Therefore,

This is our gradient.

Then we need to find y at the point x=5.

Substitute into

Now we can use the equation of

Therefore:

Expand and rearrange:

2y=x+25.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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