Graphing Quadratic Function

Question

4 (x - 9/2) -8

I can easily graph the vertex at (4.5, -8)

My question pertains to finding the parabola.

The instruction I received from IXL says:

"Now look for another point on the parabola with integer or half-integer coordinates." (This is where I am confused). "One such point is (5,-7)". (Is this point an arbitrary point, or how to you determine this point from the information given?)

Thanks,

Brian

Answer 1

Pick any value for x and substitute to determine the corresponding y coordinate. Since x = 5 was suggested here is how y = -7 was determined:

y = 4(5 - 9/2)² - 8 = 4 (1/2)² - 8 = 4(1/4) - 8 = 1 - 8 = -7.

Try x = 1/2 for another point:

y = 4( -4)² - 8 = 64 - 8 = 56, so (1/2, 56) is a point on the parabola.

You could also let y = 0 and solve for x to determine the roots of the function (x - intercepts) if any.

0 - 4(x - 9/2)² - 8

4( x - 9/2)² = 8

(x - 9/2)² = 2

x-9/2 = ±√2

x = 9/2 ±√2 (these are the x coordinates of the x intercepts of the parabola)

Visit:

desmos.com/calculator/l1hdbat8hm

to see the parabola, axis of symmetry, vertex, roots, and a couple other points on the parabola.