identify the vertex of the parabola with the quadratic equat

identify the vertex of the parabola with the quadratic equation y = 6x^2 - 72x + 220

Solution

Given quadratic equation y = 6x² - 72x + 220

General vertex form of the parabola is y=a(xh)²+k

where a-costant

(h,k)-Vertex of parabola

So,the given equation should be converted into Vertex form.

y = 6x² - 72x + 220

y= 6(x²-12x+36)+4

y=6(x²-12x+6²)+4

y=6(x²-2(x)(6)+6²)+4

y=6(x-6)²+4

By Comparing with y=a(xh)²+k

a=6 and (h,k)=(6,4)

Therefore, Vertex of the given parabola is(h,k)=(6,4)