Question

Write the standard form of the quadratic function whose graph is a parabola with the given vertex and that passes through the given point. (Let x be the independent variable and y be the dependent variable.)
Vertex: (−3, 4); point: (0, 13)

Answer 1

Answer:

The formula for this quadratic function is x*2 +6x+13

Step-by-step explanation:

If we have the vertex and one point of a parabola it is possible to find the quadratic function by the use of this

y= a (x-h)*2 + K

Quadratic function looks like this

y= ax*2 + bx + c

So let's find the a

y= a (x-h)*2 + K where

y is 13, x is 0, h is -3 and K is 4

13= a (0-(-3))*2 +4

13=9a +4

9=9a

9/9=a

1=a

The quadratic function will be

y= 1(x+3)*2 + 4

Let's get the classic form

(x+3)*2 = (x+3)(x+3)

(x*2+3x+3x+9)

x*2 +6x+13

f(0) = 13

Answer 2

Answer:

y = x^2 + 6x + 13.

Step-by-step explanation:

The general vertex  form is

y = = a(x - b)^2 + c   where a is a constant and (b, c) is the vertex.

So as the vertex is (-3, 4) we have:

y =  a(x - -3)^2 + 4

y = a(x + 3)^2 + 4

Now we find the value of a by substituting the point (0, 13)

13 = a (0 + 3)^2 + 4

13 = 9a + 4

9a = 13-4 = 9

a = 1.

So our equation is y = (x + 3)^2 + 4

Converting to Standard Form:

y = x^2 + 6x + 9 + 4

y = x^2 + 6x + 13.