Write the equation of each parabola in vertex form, which means to write in the form y= a(x-p)^2+q.?

Question

a. vertex is (1, -4) and y- intercept is 2.

b. vertex is (-2,3) congruent to y= 2x^2, and opens down.

c.vertex is (0,2) and passes through the point (-3,11)

PLEASE EXPLAIN!!

Answer 1

(a)

The vertex form of a quadratic equation is y = a(x - p)² + q
where (p, q) are the x- and y-coordinates of the vertex

vertex is (1, -4) and y- intercept is 2.

vertex (1, -4) is lies on parabola.So substitute p = 1 , q = -4
y = a(x - (-4))² + 1

y = a(x + 4)² + 1
y-intercept is at point (0, 2).
2 = a(0 + 4)² + 1

16a  = 2 - 1
a = 1/16
Therefore, the quadratic equation is: y = (1/16)(x + 4)² + 1

Answer 2

(b)

The vertex form of a quadratic equation is y = a(x - p)² + q
where (p, q) are the x- and y-coordinates of the vertex

Given that vertex is (-2,3)

So p = -2 , q = 3

Given that congruent to y= 2x² and opens down.

congruent means has the same a value (same shape) of y= 2x²

So a = 2,

But it opens down.So a = -2.

Substitute a = -2 , p = -2 , q = 3 in vortex form

y = (-2)(x - (-2))² + 3

y = -2(x + 2)² + 3

The solution is : y = -2(x + 2)² + 3

Answer 3

(c)

The vertex form of a quadratic equation is y = a(x - p)² + q
where (p, q) are the x- and y-coordinates of the vertex

vertex is (0,2) and passes through the point (-3,11)

vertex (0, 2) is lies on parabola.So substitute p = 0 , q = 2
y = a(x - 0)² + 2

y = ax² + 2
passes through the point (-3,11)
11 = a(-3)² + 2

11 - 2 = 9a

a = 9/9
a = 1
Therefore, the quadratic equation is:

y = (1)(x + 0)² + 2

y = x² + 2