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)

The standard form of the quadratic function whose graph is a parabola with the given vertex and that passes through the given point is;

y = x² + 6x + 13

We are given;

Vertex coordinate; (-3, 4)

A point on the graph; (0, 13)

The vertex form of a quadratic equation is given by;

y = a(x - h)² + k

Where h, k are the coordinates of the vertex.

a is the letter in general form of quadratic equation which is;

y = ax² + bx + c

Thus, at point (0, 13) at the vertex of (-3, 4), we have;

13 = a(0 - (-3))² + 4

⇒ 13 - 4 = 9a

9a = 9  

a = 9/9

a = 1  

Since y = a(x - h)² + k is the vertex form, let us put the vertex values for h and k as well as the value of a to get the quadratic equation;

y = 1(x - (-3))² + 4

y = x² + 6x + 9 + 4

y = x² + 6x + 13

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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

4

step-by-step explanation:

is there any photos or whatever?