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slamgirl [31]
3 years ago
8

Which equation in point-slope form contains the point (−3, 5) and has a slope of −4?

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
3 0

Answer:

y - 5 = -4(x + 3)

Step-by-step explanation:

This question is asking you to use and make an equation using the base of the "point-slope form." This is a common equation used when dealing with coordinates and graphs in math. The point-slope form equation looks like this:

y - y₁ = m(x - x₁).

We are going to need to use this equation base to create our problem from the information given. If you are wondering what those subscripts of 1 mean (the 1 in y₁ and x₁), I will explain. Remember that:

slope (m) = <u>y - y₁</u>

                  x - x₁

So, our first y value (which is the y-coordinate of 5 in [-3, 5]) can be added into the problem base that I had mentioned above:

y - <u>5</u> = m(x - x₁).

Now, we need to place the first x value (which is the -3 in [-3, 5]) can be added into the base problem once more:

y - 5 = m(x - (<u>-3</u>)).

Because a negative number with a negative symbol in front of it creates a positive, we can change that as well:

y - 5 = m(x + 3).

Fortunately, the question provides a slope ready for use. The question says that the slope is -4, so we can place this into the equation now:

y - 5 = -4(x + 3).

I hope that this helps.

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Given:

f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6

Required:

We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.

Explanation:

Given equation is

F(x)=\sqrt[4]{f(x)g(x)}.F(x)=(f(x)g(x))^{\frac{1}{4}}F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}

Differentiate the given equation for x.

Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{  Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.

F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)

Replace x=3 in the equation.

F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}F^{\prime}(3)=\frac{-6-1}{4}F^{\prime}(3)=\frac{-7}{4}

Final answer:

F^{\prime}(3)=\frac{-7}{4}

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