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 given the vertex and a point, you can use the vertex form of the equation:
y = a(x - h)^2 + k
where (h, k) represents the coordinates of the vertex.
Step 1: Use the given values for the vertex to determine the values of h and k:
h = 3
k = -4
Now we have:
y = a(x - 3)^2 - 4
Step 2: Substitute the coordinates of the given point (4,1) into the equation and solve for a:
1 = a(4 - 3)^2 - 4
1 = a(1)^2 - 4
1 = a - 4
a - 4 = 1
a = 5
Now we have the equation:
y = 5(x - 3)^2 - 4
To rewrite this equation in general form, we expand and simplify the equation:
y = 5(x - 3)(x - 3) - 4
y = 5(x^2 - 6x + 9) - 4
y = 5x^2 - 30x + 45 - 4
y = 5x^2 - 30x + 41
Therefore, the equation of the quadratic function in standard form is y = 5x^2 - 30x + 41.
To find the equation of the quadratic function in standard form, we can use the vertex form of a quadratic function, which is given by:
f(x) = a(x - h)^2 + k
Where (h, k) represents the coordinates of the vertex. In this case, the vertex is (3, -4). Plugging in the vertex values, we have:
f(x) = a(x - 3)^2 - 4
Now, we need to determine the value of 'a'. To find 'a', we use the additional point (4, 1) that the graph passes through. Substituting the point's coordinates into the equation, we get:
1 = a(4 - 3)^2 - 4
1 = a(1)^2 - 4
1 = a - 4
a = 5
Now we can rewrite the equation with the value of 'a':
f(x) = 5(x - 3)^2 - 4
To convert the equation to general form, we can expand and simplify it:
f(x) = 5(x^2 - 6x + 9) - 4
f(x) = 5x^2 - 30x + 45 - 4
f(x) = 5x^2 - 30x + 41
Thus, the equation of the function in standard form is f(x) = 5(x - 3)^2 - 4 and in general form is f(x) = 5x^2 - 30x + 41.