Hello. Please, help me solve this:
Parabola y = ax^2 + bx + c passes through the point (2;-6). Find the values of a, b and c, if vertex of parabola is (4;-10).
let's find the equation in its standard form using the given information
since the vertex is (4,-10)
it must be
y = a(x-4)^2 -10
but (2,-6) lies on it
-6 = a(-2)^2 - 10
4 = 4a
a = 1
so y = (x-4)^2 - 10
y = x^2 - 8x + 16 - 10
y = x^2 - 8x + 6
so a=1, b=-8 and c=6
To find the values of a, b, and c, we can use the information given about the vertex and a point on the parabola.
First, let's start by finding the value of a. The equation of a parabola in vertex form is given by y = a(x - h)^2 + k, where (h, k) represents the vertex.
In this case, the vertex is given as (4, -10), so we have the equation: y = a(x - 4)^2 - 10.
Next, we use the fact that the parabola passes through the point (2, -6). We substitute these coordinates into our equation and solve for a:
-6 = a(2 - 4)^2 - 10
-6 = a(-2)^2 - 10
-6 = 4a - 10
4a = -6 + 10
4a = 4
a = 1
Now that we have the value of a, we can find b and c.
To find b, we can use the fact that the vertex of the parabola is (4, -10). The x-coordinate of the vertex is given by x = -b/2a.
Substituting the values x = 4 and a = 1, we can solve for b:
4 = -b/2
8 = -b
b = -8
Finally, to find c, we substitute the values of a and b into our original equation:
y = ax^2 + bx + c
-6 = (1)(2)^2 + (-8)(2) + c
-6 = 4 - 16 + c
-6 = -12 + c
c = 6
Therefore, the values of a, b, and c are 1, -8, and 6, respectively.