Question

A quadratic function has an axis of symmetry at x=4 and contains the points (2, -6) and (7, -11). Find the equation in standard form (y=ax2+bx+c). Then state the "a" value in the equation.

Answer 1

Answer:

The equation in standard form is $y = -x^{2}+8\cdot x -18$. 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:

$y - k = C\cdot (x-h)^{2}$ (1)

Where:

$C$ - Parabola constant, dimensionless.

$x$ - Independent variable, dimensionless.

$y$ - Depedent variable, dimensionless.

$h$, $k$ - Coordinates of the vertex of the parabola, dimensionless.

If we know that $(x_{1},y_{1})=(2,-6)$, $(x_{2},y_{2})=(7,-11)$ and $h = 4$, then we have the following system of linear equations:

$(x_{1},y_{1})=(2,-6)$

$-6-k = C\cdot (2-4)^{2}$

$4\cdot C +k = -6$ (2)

$(x_{2},y_{2})=(7,-11)$

$-11-k =C\cdot (7-4)^{2}$

$9\cdot C + k = -11$ (3)

By clearing $k$ in (2) and (3) and equalizing each other, we get that:

$-6-4\cdot C = -11-9\cdot C$

$5\cdot C = -5$

$C = -1$

And the remaining variable is calculated by substituting directly on (3):

$k = -11-9\cdot C$

$k = -11-9\cdot (-1)$

$k = -2$

Then, the equation of the parabola is:

$y+2 = -(x-4)^{2}$

And the standard form of the equation is obtained by algebraic handling:

$y+2 = -(x^{2}-8\cdot x + 16)$

$y + 2 = -x^{2}+8\cdot x -16$

$y = -x^{2}+8\cdot x -18$ (4)

The equation in standard form is $y = -x^{2}+8\cdot x -18$. The "a" value is -1.