Question

The graph of the quadratic $y = ax^2 + bx + c$ is a parabola that passes through the points $(-1,7)$, $(5,7)$, and $(6,10)$. What is the $x$-coordinate of the vertex of the parabola?

Answer 1

Answer:

$V(x,y) = \left(3,\frac{23}{5} \right)$

Step-by-step explanation:

Let consider the following linear equation systems by using the known points and second-grade polynomial:

(-1, 7)

$a + b + c = 7$

(5, 7)

$25\cdot a + 5\cdot b + c = 7$

(6,10)

$36\cdot a + 6 \cdot b + c = 10$

After some algebraic manipulation, the values for the polynomial coefficients are found:

$a = \frac{3}{5}$, $b = -\frac{18}{5}$, $c = 10$

The polynomial is:

$y = \frac{3}{5}\cdot x^{2} - \frac{18}{5}\cdot x +10$

Lastly, the vertex is found by further handling:

$y - 10 = \frac{3}{5}\cdot (x^{2}-6\cdot x + 9) - \frac{27}{5}$

$y -\frac{23}{5} = \frac{3}{5}\cdot (x-3)^{2}$

The coordinates for the vertex are:

$V(x,y) = \left(3,\frac{23}{5} \right)$

Answer 2

Two of the given points have the same y-value. The midpoint of those two will be on the line of symmetry, as is the vertex. The x-value there is (-1+5)/2 = 2.