Question

Find the vertex for the parabola whose equation is given by writing the equation in the form y = ax² +bx+c.y=(x-3)²-8(Type an ordered pair.)The vertex is

Answer 1

Given that

The equation of a parabola is y = (x-3)²-8

and we have to find its vertex.

Explanation -

First, we will write the given equation in the form of y = ax² +bx+c---------(i)

Then, we will use the formula

$(a-b)^2=a^2+b^2+2ab$

y = (x-3)²-8

y = x² + 9 - 6x - 8

y = x² - 6x + 1-------------(ii)

Comparing equation (i) and (ii) we have

a = 1, b = -6 and c = 1

and from this equation we can find the vertex and the formula for vertex is

$Vertex=(\frac{-b}{2a},\frac{-(b^2-4ac)}{4a})$

On substituting the values we have

$Vertex=(\frac{-(-6)}{2},\frac{-(36-4\times1\times1}{4})=(\frac{6}{2},\frac{-(36-4)}{4})=(3,\frac{-32}{4})=(3,-8)$

The coordinates of the vertex is (3,-8)

Hence, the final answer is (3,-8)