Question

Which equation represents a parabola that opens upward has a minimum at x=3 and has a line of symmetry at x=3

Answer 1

Answer:

$y=x^2-6x+5$

Step-by-step explanation:

Let us consider the equation $y=x^2-6x+5$

For a quadratic equation in a standard form, $y=ax^2+bx+c$, the axis of symmetry is the vertical line $x = \frac{-b}{2a}$.

Here in this case we have, $a=1, b=-6 , c =5$

Putting the values we get,

$x = \frac{-(-6)}{2\times 1} = \frac{6}{2} =3$

We can see that the axis of symmetry is x=3 and the graph is giving minimum at x=3.

Therefore, the required equation is $y=x^2-6x+5$. Refer the image attached.

Answer 2

Answer:

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

Step-by-step explanation: