determine the quadratic function f whose graph is given.
the vertex is (1,-3)
y intercept is -2
please show work so i can see how this is done, thank you!
Find the equation of the quadratic Functions whose graph is a parabola the points (-2,19), (3,14) and (4,25)
To determine the quadratic function f, we can use the vertex form of a quadratic equation:
f(x) = a(x - h)^2 + k
Where (h, k) represents the vertex. In this case, the vertex is (1, -3). Hence, the equation becomes:
f(x) = a(x - 1)^2 - 3
Now, we need to find the value of "a". To do this, we can use the y-intercept, which is (-2, 0). We substitute these values into the equation:
0 = a(-2 - 1)^2 - 3
0 = a(-3)^2 - 3
0 = a(9) - 3
0 = 9a - 3
Now, solve for "a" by isolating it:
9a = 3
a = 3/9
a = 1/3
Substitute the value of "a" back into the equation:
f(x) = (1/3)(x - 1)^2 - 3
Therefore, the quadratic function f whose graph is given with a vertex at (1, -3) and y-intercept at -2 is:
f(x) = (1/3)(x - 1)^2 - 3
To determine the quadratic function f given the vertex and y-intercept, we can use the standard form of a quadratic equation:
f(x) = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola and 'a' determines the direction and width of the parabola.
Given that the vertex is (1, -3), we can substitute these values into the equation:
f(x) = a(x - 1)^2 - 3
Now, let's use the y-intercept to find the value of 'a.'
The y-intercept occurs when x = 0, and the y-coordinate is -2. So we have:
f(0) = a(0 - 1)^2 - 3
-2 = a(1) - 3
-2 + 3 = a
a = 1
With the value of 'a,' we can rewrite the equation as:
f(x) = (x - 1)^2 - 3
So, the quadratic function f(x) with the given vertex (1,-3) and y-intercept -2 is:
f(x) = (x - 1)^2 - 3
For a vertex of (1,-3) the equation is
Y= a(x-1)^2 - 3
But (0,-2)lies on it, so
-2 = a(0-1)^2 -3
1 = a
Y = (x-1)^2 - 3