Someone pls help. Write the equation in standard form

Step-by-step explanation:

Since, zeros of the relation are 5, 6 and y-intercept is (0, 30).

Therefore, equation of the quadratic relation will be in the form of,

y = a(x - 5)(x - 6)

For (0, 30),

30 = a(0 - 5)(0 - 6)

30 = 30a

a = 1

Therefore, equation will be,

y = (x - 5)(x - 6)

y = x² - 11x + 30

y = x²- 2(5.5)x + (5.5)² - (5.5)² + 30

y = (x - 5.5)²+ 30 - 30.25

y = (x - 5.5)² - 0.25 [Standard form}  

b). Vertex → (-4, -7) and passes through a point (-2, 1)

Vertex form or standard equation of the quadratic relation is,

y = a(x - h)² + k

Here, (h, k) is the vertex

Therefore, equation will be,

y = a(x + 4)² - 7

Since, (-2, 1) lies on this relation

1 = a(-2 + 4)² - 7

a(2)²- 7 = 1

4a = 8

a = 2

Equation will be,

y = 2(x + 4)²- 7