Question

There exist constants a,h,k and such that 3x^2+12x+4=a(x-h)^2+k for all real numbers x Enter the ordered triple (a,h,k) Find the vertex of the graph of the equation x-y^2+8y=13 find the vertex of the graph of the equation y=-2x^2+8x-15

Answer 1

Given :  There exist constants a,h,k and such that 3x^2+12x+4=a(x-h)^2+k for all real numbers

To find : Find the ordered triple (a,h,k)

Solution:

3x²+12x+4=a(x-h)²+k

3x²+12x+4

=  3(x² + 4x)  + 4

=  3(x² + 4x)  + 4  + 12 - 12

= 3(x² + 4x) + 12  - 8

=3(x² + 4x + 4) - 8

= 3(x + 2)² - 8

= 3(x -(-2))² - 8

Comparing with

a(x-h)²+k

=> a = 3   ,  h = - 2  , k = - 8

( 3 , -2 , - 8)

(-2 , - 8) is vertex

x-y²+8y=13

=>x = y² - 8y + 13

=> x = (y - 4)²  - 16 + 13

=> x = (y - 4)²  - 3

y = 4  , x  = - 3

(-3 , 4) is Vertex

y=-2x²+8x-15

=> y = -2(x² - 4x  + 4) + 8 - 15

=> y = -2(x - 2)² - 7

x = 2,     y = -7

(2 , - 7) is vertex

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