Question

Y=x^2-6x+3 help write the vertex form of the equation

Answer 1

Answer:

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

Step-by-step explanation:

$y=x^2-6x+3$

This is written in the standard form of a quadratic function:

$y=ax^2+bx+c$

where:

• ax² → quadratic term
• bx → linear term
• c → constant

You need to convert this to vertex form:

$y=a(x-h)^2+k$

where:

• (h,k) → vertex

To find the vertex form, you need to find the vertex. For this, use the equation for axis of symmetry, since this line passes through the vertex:

$x=-\frac{b}{2a}$

Using your original equation, identify the a, b, and c terms:

$a=1\\\\b=-6\\\\c=3$

Insert the known values into the equation:

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

Simplify. Two negatives make a positive:

$x=\frac{6}{2} =3$

X is equal to 3 (3,y). Insert the value of x into the standard form equation and solve for y:

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

Simplify using PEMDAS:

$y=9-18+3\\\\y=-9+3\\\\y=-6$

The value of y is -6 (3,-6). Insert these values into the vertex form:

$(3_{h},-6_{k})\\\\y=a(x-3)^2+(-6)$

Insert the value of a and simplify:

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

:Done