3 years ago

Enter the values of h and I so that y=x2+6x+10 is in vertex form Y=(x+)^2+_

1 answer:

3 0

$Use\ (a+b)^2=a^2+2ab+b^2\qquad(*)\\\\y=x^2+6x+10=x^2+2(x)(3)+10=\underbrace{x^2+2(x)(3)+3^2}_{(*)}-3^2+10\\\\=(x+3)^2-9+10=(x+3)^2+1\\\\Answer:\ \boxed{y=(x+3)^2+1}$

You might be interested in

Perform the indicated operation 1. (12 + 3) - 7 = N

lubasha [3.4K]

Answer:

N = 8

Step-by-step explanation:

(12 + 3) - 7 = N

=> N = (12 + 3) - 7

=> N = 15 - 7

=> N = 8

8 0

2 years ago

(x2 + 4x2 + x3 + x) + (x3 + x + 5x + 5)

yuradex [85]

Answer:

2x³+5x²+7x+5

Step-by-step explanation:

(x² + 4x² + x³ + x) + (x³ + x + 5x + 5)

(5x²+x³+x)+(x³+6x+5)

2x³+5x²+7x+5

3 0

3 years ago

If m < 0 and b > 0, the graph of y = mx + b does not pass through which quadrant?

stiks02 [169]

Answer:

Quadrant III

Step-by-step explanation:

The attached picture shows graph of 4 such linear functions with the conditions given in the problem. ALL of them DO NOT pass through Quadrant III.

The graphs shown are of the functions:

$y=-2x+1$