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Setler [38]
4 years ago
14

The following graph represents the hourly commission of a home appliance salesperson, y, based on the number of salespeople that

are
working the showroom floor, x. The graph of the hourly commission of the salesperson is a quadratic function. What is the equation for
the function in standard form if the vertex is 6.5,42.25 and the origin is the only other known point?

Mathematics
1 answer:
grandymaker [24]4 years ago
3 0

We have been given graph of a downward opening parabola with vertex at point (6.5,42.25). We are asked to write equation of the parabola in standard form.

We know that equation of parabola in standard form is f(x)=ax^2+bx+c.

We will write our equation in vertex form and then convert it into standard form.

Vertex for of parabola is y=a(x-h)^2+k, where point (h,k) represents vertex of parabola and a represents leading coefficient.

Since our parabola is downward opening so leading coefficient will be negative.

Upon substituting coordinates of vertex and point (0,0) in vertex form, we will get:

0=-a(0-6.5)^2+42.25

0=-a(42.25)+42.25

-42.25=-a(42.25)+42.25-42.25

-42.25=-42.25a

a=1 Divide both sides by -42.25

So our equation in vertex form would be f(x)=-(x-6.5)^2+42.25.

Let us convert it in standard from.

f(x)=-(x^2-13x+42.25)+42.25

f(x)=-x^2+13x-42.25+42.25

f(x)=-x^2+13x

Therefore, the equation of function is standard form would be f(x)=-x^2+13x.

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Step-by-step explanation:

First factor out the negative sign from the expression and reorder the terms

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )}  -  \frac{1}{ \cot(6A)  -  \cot(2A) }

<u>Using trigonometric </u><u>identities</u>

That's

<h3>\cot(x)  =  \frac{1}{ \tan(x) }</h3>

<u>Rewrite the expression</u>

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )} -    \frac{1}{ \frac{1}{ \tan(6A) } }  -  \frac{1}{ \frac{1}{ \tan(2A) } }

We have

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{1}{ \frac{ \tan(2A) -  \tan(6A)  }{ \tan(6A) \tan(2A)  } }</h3>

<u>Rewrite the second fraction</u>

That's

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{ \tan(6A)  \tan(2A) }{ \tan(2A) -  \tan(6A)  }</h3>

Since they have the same denominator we can write the fraction as

-  \frac{1 +  \tan(6A) \tan(2A)  }{ \tan(2A) -  \tan(6A)  }

Using the identity

<h3>\frac{x}{y}  =  \frac{1}{ \frac{y}{x} }</h3>

<u>Rewrite the expression</u>

We have

<h3>-  \frac{1}{ \frac{ \tan(2A)  -  \tan(6A) }{1 +  \tan(6A) \tan(2A)  } }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{ \tan(x) -  \tan(y)  }{1 +  \tan(x)  \tan(y) }  =  \tan(x - y)</h3>

<u>Rewrite the expression</u>

That's

<h3>- \frac{1}{ \tan(2A -6A) }</h3>

Which is

<h3>-  \frac{1}{ \tan( - 4A) }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{1}{ \tan(x) }  =  \cot(x)</h3>

Rewrite the expression

That's

<h3>-  \cot( - 4A)</h3>

<u>Simplify the expression using symmetry of trigonometric functions</u>

That's

<h3>- ( -  \cot(4A) )</h3>

<u>Remove the parenthesis </u>

We have the final answer as

<h2>\cot(4A)</h2>

As proven

Hope this helps you

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