Answer:
The equation in standard form is [tex]y = -x^{2}+8\cdot x -18[/tex]. The "a" value is -1.
Step-by-step explanation:
A quadratic function is the standard form of the parabola. We can take advantage of the symmetry property of the parabola by using the following formula from the Analytical Geometry:
[tex]y - k = C\cdot (x-h)^{2}[/tex] (1)
Where:
[tex]C[/tex] - Parabola constant, dimensionless.
[tex]x[/tex] - Independent variable, dimensionless.
[tex]y[/tex] - Depedent variable, dimensionless.
[tex]h[/tex], [tex]k[/tex] - Coordinates of the vertex of the parabola, dimensionless.
If we know that [tex](x_{1},y_{1})=(2,-6)[/tex], [tex](x_{2},y_{2})=(7,-11)[/tex] and [tex]h = 4[/tex], then we have the following system of linear equations:
[tex](x_{1},y_{1})=(2,-6)[/tex]
[tex]-6-k = C\cdot (2-4)^{2}[/tex]
[tex]4\cdot C +k = -6[/tex] (2)
[tex](x_{2},y_{2})=(7,-11)[/tex]
[tex]-11-k =C\cdot (7-4)^{2}[/tex]
[tex]9\cdot C + k = -11[/tex] (3)
By clearing [tex]k[/tex] in (2) and (3) and equalizing each other, we get that:
[tex]-6-4\cdot C = -11-9\cdot C[/tex]
[tex]5\cdot C = -5[/tex]
[tex]C = -1[/tex]
And the remaining variable is calculated by substituting directly on (3):
[tex]k = -11-9\cdot C[/tex]
[tex]k = -11-9\cdot (-1)[/tex]
[tex]k = -2[/tex]
Then, the equation of the parabola is:
[tex]y+2 = -(x-4)^{2}[/tex]
And the standard form of the equation is obtained by algebraic handling:
[tex]y+2 = -(x^{2}-8\cdot x + 16)[/tex]
[tex]y + 2 = -x^{2}+8\cdot x -16[/tex]
[tex]y = -x^{2}+8\cdot x -18[/tex] (4)
The equation in standard form is [tex]y = -x^{2}+8\cdot x -18[/tex]. The "a" value is -1.
