Question

Write a quadratic equation in standard form that has a vertex of (-2, 6) and passes through (-4, -2).

Answer 1

Answer:

The equation of the quadratic in standard form is:

$y=-2x^2-8x-2$

Step-by-step explanation:

Since they give us the information about where the vertex of the parabola is located, and one extra points where it passes through, we can use the general form of a quadratic in vertex form:

$y-y_v=a\,(x-x_v)^2$

where $(x_v,y_v)$ is the location of the vertex (in our case the point (-2,6).

Therefore the equation above becomes:

$y-y_v=a\,(x-x_v)^2\\y-6=a\,(x-(-2))^2\\y-6=a\,(x+2)^2$

Now,we can use the fact that the point (-4,-2) is also a point of the graph, to find the value of the parameter $"a"$:

$y-6=a\,(x+2)^2\\-2-6=a\,(-4+2)^2\\-8=a\,(-2)^2\\-8=a\,*4\\a=-2$

Then, the equation of the quadratic with such characteristics becomes:

$y-6=-2\,(x+2)^2\\y-6=-2x^2-8x-8\\y=-2x^2-8x-2$

which is the equation of the quadratic in standard form.