Question

Which equation represents a parabola that opens upward, has a minimum at x = 3, and has a line of symmetry at x = 3?
A. y = x^2 -6x + 13

B. y = x^2 + 6x + 5

C. y = x^2 - 3x + 6

D. y = x^2 + 8x + 19

Answer 1

Answer:

A. y = x^2 -6x + 13

Step-by-step explanation:

The parabola that opens up and has line of symmetry x=3 and a minimum at x=3  is

$y=x^2-6x+13$

The reason is that, we can write this function in the vertex form to get;

$y=x^2-6x+(-3)^2+13-(-3)^2$

$y=(x-3)^2+13-9$

$y=(x-3)^2+4$

Hence the line of symmetry is x=3 and the vertex is (3,4)

Answer 2

Answer:

option A

y = x² - 6x + 5

Step-by-step explanation:

step 1

Find out if a is positive or negative.

'a' is the coefficent of x

If the parabola is facing up , then a must be positive.

a = 1

Step 2

To find the minimum point

x = -b/2a

using the standard equation y = ax² + bx + c

x must be equal to 3

Equation 1

y = x² -6x + 13

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

x =  3(ACCEPTED)

Equation 2

y = x² + 6x + 5

x = -6/2(1)

x = -3      

Equation 3

x² - 3x + 6

x =  3/2(1)

x =  3/2

Equation 4

x² + 8x + 19

x = -8/2(1)

x = -4

Step 3

At the minimum point there will be line of symmetry hence x = 3