Question

What is the vertex of the function x2 - 6x + 8 = 0?

Answer 1

Answer:

The vertex of the function $\bold{x^{2} -6 x + 8 = 0}$is (h,k) = (3 , -1)

Solution:

The vertex form of quadratic equation is generally given as,

$f(x) = a(x - h)^{2} + k$

Where h,k is the vertex of the parabola.

From question, given that $x^{2}-6 x+8=0$ .

we have to find the vertex of the function.

Let us first convert the given quadratic equation to vertex form (eqn 1)

$x^{2}-6 x+8=0$

$x^{2}-6 x=-8$

By adding “9” on both sides of equation, we get

$x^{2}-6 x+9=-8+9$

$x^{2}-6 x+9=1$

By using the identity $(a-b)^{2}=a^{2}-2 a b+b^{2}$ ,the right hand side of above equation becomes,

$(x-3)^{2}=1$

$(x-3)^{2}-1=0$

Now,the equation $(x-3)^{2}-1=0$ is of the vertex form.

By comparing $(x-3)^{2}-1=0$ with $a(x-h)^{2}+k$

we get the values of (h,k)

a = 1; h = 3; k = -1

hence the vertex of the function $x^{2}-6 x+8=0$ is (h,k) = (3 , -1)