The graph of a quadratic function has vertex(3,-4) and passes through the point (4,1).

Find the equation of the function in standard form.

Rewrite the equation in general form.

Could someone explain to me how to do this, just don't understand.

To find the equation of a quadratic function in standard form using the given vertex and a point, you can use the vertex form of the equation, which is:

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

where (h, k) represents the vertex.

Step 1: Use the vertex coordinates to determine the values of h and k.
Given vertex (3, -4):
h = 3
k = -4

Step 2: Use the point (4, 1) to find the value of a.
Plug in the point's coordinates into the vertex form equation and solve for a:
1 = a(4 - 3)^2 - 4
1 = a(1)^2 - 4
1 = a - 4
a = 1 + 4
a = 5

Therefore, the equation in vertex form is:
f(x) = 5(x - 3)^2 - 4

To rewrite this equation in general form, expand and simplify:

f(x) = 5(x^2 - 6x + 9) - 4
f(x) = 5x^2 - 30x + 45 - 4
f(x) = 5x^2 - 30x + 41

So, the equation in general form is:
5x^2 - 30x + 41

To find the equation of a quadratic function given the vertex and a point on the graph, you can use the vertex form of the equation. The vertex form equation of a quadratic function is y = a(x - h)^2 + k, where (h, k) represents the vertex coordinates.

1. Find the value of "a" using the given point (4,1):
Substituting the vertex coordinates (3, -4) into the vertex form equation, we get -4 = a(3 - 3)^2 + (-4), which simplifies to -4 = a(0) - 4. Therefore, "a" = -4.

2. Now that we know the value of "a", we can plug it back into the vertex form equation:
y = -4(x - 3)^2 + (-4)
Simplifying further, we get y = -4(x - 3)^2 - 4.

3. Finally, we can rewrite the equation in standard form, which is in the form of y = ax^2 + bx + c:
Expanding the equation, we have y = -4(x^2 - 6x + 9) - 4. Distributing -4, we get y = -4x^2 + 24x - 36 - 4. Combining like terms, the equation can be presented as y = -4x^2 + 24x - 40.

So, the equation of the quadratic function in standard form is y = -4x^2 + 24x - 40.

The canonical form of a quadratic equation is:

y=a(x-h)²+k
where
a is a constant
(h,k) is the vertex.

So in the given case, h=3, k=-4, and a remains to be found.
We can find a since we know the curves passes through (4,1), hence
1=a(4-3)²-4
1=a-4
solve for a to get
a=5
Check:
y=5(x-3)²-4
For x=4, y=5(1)²-4=1, OK

For the general form, expand the equation
y=5(x-3)²-4
y=5x²-30x+41