Write the standard form of the quadratic function that has the indicated vertex and whose graph passes through the given point.

Vertex: (-2, 5)
Point: (0, 9)

Well, well, well, it looks like we've got ourselves a quadratic function conundrum! To find the standard form of this quadratic function, we're gonna need to gather some intel.

First, let's find the equation in vertex form. The vertex form of a quadratic function is given by:

f(x) = a(x - h)^2 + k

Where (h, k) represents the vertex. In this case, the vertex is (-2, 5), so we can plug these values into our equation:

f(x) = a(x - (-2))^2 + 5
f(x) = a(x + 2)^2 + 5

Now we need to find the value of 'a'. We can do this by using the given point (0, 9) and substituting it into our derived equation. Here's the calculation:

9 = a(0 + 2)^2 + 5
9 = a(2)^2 + 5
9 = 4a + 5
4a = 9 - 5
4a = 4
a = 1

Finally, we can substitute the value of 'a' we just found back into our equation:

f(x) = 1(x + 2)^2 + 5

Voila! The standard form of the quadratic function that satisfies the given vertex and passes through the point (0, 9) is:

f(x) = (x + 2)^2 + 5

To write the standard form of the quadratic function that has the given vertex and passes through the given point, we can use the vertex form of a quadratic function, which is:

f(x) = a(x - h)^2 + k

where (h, k) represents the vertex. In this case, the vertex is (-2, 5), so h = -2 and k = 5. Substituting these values into the vertex form, we have:

f(x) = a(x - (-2))^2 + 5
f(x) = a(x + 2)^2 + 5

To find the value of "a," we substitute the given point (0, 9) into the equation:

9 = a(0 + 2)^2 + 5
9 = a(2)^2 + 5
9 = 4a + 5
4a = 9 - 5
4a = 4
a = 4/4
a = 1

Therefore, the standard form of the quadratic function is:

f(x) = (x + 2)^2 + 5

To find the standard form of a quadratic function, we need to consider the vertex form of a quadratic function and then convert it to standard form. The vertex form of a quadratic function is given by:

f(x) = a(x - h)^2 + k,

where (h, k) is the vertex of the parabola.

Given that the vertex is (-2, 5), we can substitute these values into the vertex form:

f(x) = a(x - (-2))^2 + 5.

Simplifying further, we have:

f(x) = a(x + 2)^2 + 5.

Now we need to use the given point (0, 9) to find the value of 'a'. Substitute the values of x and f(x) into the equation:

9 = a(0 + 2)^2 + 5.

Simplifying:

9 = a(2)^2 + 5
9 = 4a + 5
4 = 4a
a = 1.

Now that we have the value of 'a', we can substitute it back into the vertex form:

f(x) = 1(x + 2)^2 + 5.

Finally, let's expand and simplify the equation to get the standard form:

f(x) = (x + 2)^2 + 5
f(x) = x^2 + 4x + 4 + 5
f(x) = x^2 + 4x + 9.

Therefore, the standard form of the quadratic function with the given vertex and point is f(x) = x^2 + 4x + 9.

Standard quadratic function with given vertex (h,k) is given by:

y=a(x-h)²+k
where a can be found by substituting the values of any given point on the curve.
Here, h=-2, k=5, so
y=a(x+2)²+5
Now substitute x=0, y=9 and solve for a.