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Please check the photo (Calculus), I just need the first part of this answer solved.

Please check the photo (Calculus), I just need the first part of this answer solved-example-1
User Jan Spurny
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The given function is


f(x)=15-x^2^{}

First, we have to find f(x+h), which consists of adding h to the x-variable.


f(x+h)=15-(x+h)^2

Then, we use the functions we have in the slope definition.


\lim _(h\to0)((f(x+h)-f(x))/(h))=\lim _(h\to0)((15-(x+h)^2-(15-x^2))/(h))

Solve the notable product and multiply by the negative sign in front.


\lim _(h\to0)((15-(x^2+2xh+h^2)-15+x^2)/(h))=\lim _(h\to0)((15-x^2-2xh-h^2-15+x^2)/(h))

Add like terms.


\lim _(h\to0)((-2xh-h^2)/(h))

Factor out the greatest common factor h.


\lim _(h\to0)(h(-2x-h))/(h)=\lim _(h\to0)(-2x-h)

At last, evaluate the expression when h = 0.


\lim _(h\to0)(-2x-h)=-2x-0=-2x

The slope is given by the equation -2x.

Let's evaluate the expression when x = 2 to find the value of the slope.


m=-2x=-2(2)=-4

(a) Therefore, the slope is -4.

To find the equation for the tangent, we have to use the point-slope formula.


\begin{gathered} y-y_1=m(x-x_1);\text{ where} \\ x_1=2,y_1=11,m=-4 \end{gathered}

Use the values of the coordinates and slope to find the equation.


\begin{gathered} y-11=-4(x-2) \\ y-11=-4x+8 \\ y=-4x+8+11 \\ y=-4x+19 \end{gathered}

(b) Therefore, the equation for the tangent to the curve is y = -4x+19.

(c)

The image below shows the function f(x) and the tangent.

Please check the photo (Calculus), I just need the first part of this answer solved-example-1
User Andrew Pate
by
8.8k points

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