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katen-ka-za [31]
3 years ago
9

What are the vertex and x-intercepts of the graph of the function below?

Mathematics
1 answer:
geniusboy [140]3 years ago
5 0

y= x^2 - 6x-7

Vertex = -b /2a

b= -6

a= 1

-(-6)/2(1)= 3

x= 3

Use the substitution method to find y

y= (3)^2-6(3)-7

(3)(3)-18-7

9-25

y= -16

Vertex is (3, 16)

Find x- intercept (y= 0)

y= x^2 - 6x-7

(x-7)(x+1)

x-7= 0

Move -7 to the other side. Sign changes from -7 to 7

x-7+7= 0+7

x= 7

x+1= 0

Move +1 to the other side. Sign changes from +1 to -1

x+1-1= 0-1

x= -1

Answer :B. Vertex: (3, -16); Intercepts: x=-1,7

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Determine the corresponding general form of each of the following vertex form
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1) y = (x -1)²

 y= x² -  2*x*1 + 1

y = x² - 2x + 1    

Ans: C

2)y = (x +4)² + 5

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3) y = -(x + 9)²- 10

y = - [x² + 18x + 81] - 10

= -x² - 18x - 81 - 10

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4) y = 3(x + 2)² - 18

y =3 [x² + 4x + 4] - 18

y =  3x² + 12x  + 12 - 18

y =3x²  + 12x - 6

A

5) y = -2(x + 1)² - 16

       = -2[x² + 2x + 1] -16

       = -2x² - 4x - 2 - 16

     y = -2x² - 4x -  18

A

6) y = 5(x + 5)²

        =5[x²+ 10x + 25]

y = 5x² +50x + 125

A

7)y = (1/2)(x + 8)² - 8

  y = (1/2) (x² + 16x + 64) - 8

y =  \dfrac{1}{2}*x^{2}+ \dfrac{1}{2}*16x  + \dfrac{1}{2}*64 -8\\\\\\y =\dfrac{1}{2}x^{2}+8x +32 - 8\\\\\\y=\dfrac{1}{2}x^{2} +8x + 24

A

8) y = (x + 3/2)² + 3/4

y = x^{2} +2*x*\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{3}{4}\\\\y=x^{2}+3x+\dfrac{12}{4}\\\\\\y=x^{2}+3x + 3

C

9) y = 2[x² + 16x + 64] - 5x

y = 2x² + 32x + 64 - 5x

y =2x² + 27x + 6

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