Question

Quadratics need help making parabola

Answer 1

Answer:

The equation of the parabola is $y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4$, whose real vertex is $(x,y) = (2, -5.333)$, not $(x,y) = (2, -5)$.

Step-by-step explanation:

A parabola is a second order polynomial. By Fundamental Theorem of Algebra we know that a second order polynomial can be formed when three distinct points are known. From statement we have the following information:

$(x_{1}, y_{1}) = (-2, 0)$, $(x_{2}, y_{2}) = (6, 0)$, $(x_{3}, y_{3}) = (0, -4)$

From definition of second order polynomial and the three points described above, we have the following system of linear equations:

$4\cdot a -2\cdot b + c = 0$ (1)

$36\cdot a + 6\cdot b + c = 0$ (2)

$c = -4$ (3)

The solution of this system is: $a = \frac{1}{3}$, $b = - \frac{4}{3}$, $c = -4$. Hence, the equation of the parabola is $y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4$. Lastly, we must check if $(x,y) = (2, -5)$ belongs to the function. If we know that $x = 2$, then the value of $y$ is:

$y = \frac{1}{3}\cdot (2)^{2}-\frac{4}{3}\cdot (2) - 4$

$y = -5.333$

$(x,y) = (2, -5)$ does not belong to the function, the real point is $(x,y) = (2, -5.333)$.