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melomori [17]
3 years ago
14

What is the vertex form of the parabola whose standard form equation is y=5x^2-30x+49

Mathematics
1 answer:
agasfer [191]3 years ago
7 0

Answer:

The vertex is (3,4)

Step-by-step explanation:

To convert a quadratic from  

y = a x 2 + b x + c

form to vertex form,

y = a ( x − h ) 2 + k , you use the process of completing the square.

First, we must isolate the  x

terms:

y − 49 = 5 x 2 − 30 x + 49 − 49

y − 49 = 5 x 2 − 30 x

We need a leading coefficient of  1

for completing the square, so factor out the current leading coefficient of 2.

y − 49 = 5 ( x 2 − 6 x )

Next, we need to add the correct number to both sides of the equation to create a perfect square. However, because the number will be placed inside the parenthesis on the right side we must factor it by  

2

on the left side of the equation. This is the coefficient we factored out in the previous step.

y − 49 + ( 5 ⋅ ? ) = 5 ( x 2 − 6 x + ? )

<- Hint:  

62 = 3 ;  3 ⋅ 3 = 9

y − 49 + ( 5 ⋅ 9 ) = 5 ( x 2 − 6 x + 9 )  

y− 49 + 45 = 5 ( x 2 − 6 x + 9 )

y − 4 = 5 ( x 2 − 6 x + 9 )

Then, we need to create the square on the right hand side of the equation:

y − 4 = 5 ( x − 3 ) 2

Now, isolate the  y  term:

y − 4 + 4 = 5 ( x − 3 ) 2 + 4

y − 0 = 5 ( x − 3 ) 2 + 4

y − 0 = 5 ( x − 3 ) 2 + 4

The vertex is:  

( 3 , 4 )

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The expression can be solved by expanding the bracket and multiplying out the terms

(p^4-q^4)(p^4+q^4)\begin{gathered} =p^4(p^4+q^4)-q^4(p^4+q^4) \\ =p^8+p^4q^4-p^4q^4-q^8 \\ =p^8-q^8 \end{gathered}

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