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3 years ago

AJ graphs the function f(x)=-(x+2)^(2)-1

Mathematics

1 answer:

irina1246 [14]3 years ago

7 0

Answer:

Part 1: AJ reflected over x-axis.

Part 2: $(0,-5)$ and $(-1,-2)$

AJ plotted the graph using the function $f(x)=(x+2)^{2}-1$ . This resulted the graph to reflect over x-axis

Step-by-step explanation:

Part 1: AJ mistakenly plotted over x-axis, which means he reflected the graph over x-axis.The x-value remains the same. Only the y-values are transformed into its opposite sign.

Part 2:

Step 1: Plotting any two values for the function $f(x)=-(x+2)^{2}-1$

Substituting x=0, we get,

$\begin{aligned}y=f(0) &=-(0+2)^{2}-1 \\=&-(2)^{2}-1 \\=&-4-1=-5 \\& y=-5\end{aligned}$

Substituting x=-1, we get,

$\begin{aligned}y=f(-1) &=-(-1+2)^{2}-1 \\=&-(1)^{2}-1 \\=&-1-1=-2 \\& y=-2\end{aligned}$

The two x-values for AJ’s function is $(0,-5)$ and $(-1,-2)$

Step 2:

AJ plotted the graph using the function $f(x)=(x+2)^{2}-1$ . This resulted the graph to reflect over x-axis.

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\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
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\begin{cases}
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recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
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cos(\theta)=\cfrac{adjacent}{hypotenuse}
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% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
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% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
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sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

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tan(\theta)=\cfrac{-1}{\sqrt{35}}
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cot(\theta)=\cfrac{\sqrt{35}}{1}
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sec(\theta)=\cfrac{6}{\sqrt{35}}$

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sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

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