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?

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.

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xero099 xero099 · Answer 2 · 2020-04-22T02:34:00+02:00

Answer:

[tex]V(x,y) = \left(3,\frac{23}{5} \right)[/tex]

Step-by-step explanation:

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

(-1, 7)

[tex]a + b + c = 7[/tex]

(5, 7)

[tex]25\cdot a + 5\cdot b + c = 7[/tex]

(6,10)

[tex]36\cdot a + 6 \cdot b + c = 10[/tex]

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

[tex]a = \frac{3}{5}[/tex], [tex]b = -\frac{18}{5}[/tex], [tex]c = 10[/tex]

The polynomial is:

[tex]y = \frac{3}{5}\cdot x^{2} - \frac{18}{5}\cdot x +10[/tex]

Lastly, the vertex is found by further handling:

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

[tex]y -\frac{23}{5} = \frac{3}{5}\cdot (x-3)^{2}[/tex]

The coordinates for the vertex are:

[tex]V(x,y) = \left(3,\frac{23}{5} \right)[/tex]