+0

0
100
1

Suppose that we have an equation y = ax^2 + bx + c whose graph is a parabola with vertex (-4,7), vertical axis of symmetry, and contains the point (2,-1).  What is (a,b,c)?

Sep 22, 2020

#1
+1

Suppose that we have an equation y = ax^2 + bx + c whose graph is a parabola with vertex (-4,7), vertical axis of symmetry, and contains the point (2,-1).  What is (a,b,c)?

Hello Guest!

$$V (-4,7)\\ P_1(2,-1)$$

Because of the symmetry around $$x = -4$$, is

$$P_2(-10,-1)$$

The coordinates of the 3 points are used in equation $$y = ax^2 + bx + c$$, and the resulting equations are solved for a, b, c.

I. $$7=16a-4b+c$$

II. $$-1=4a+2b+c$$

III. $$-1=100a-10b+c$$

III. - II. $$0=96a-12b$$

I. - II. $$8=12a-6b$$

III. - II.           $$0=96a-12b$$

- 2 * (I. - II.) $$\underline{16=24a-12b}$$

$$-16=72a$$

$$a=-\frac{2}{9}$$

III. - II.          $$0=96\cdot (-\frac{2}{9})-12b\\ 0=-21.\overline{3}-12b$$

$$b=-1.\overline{7}$$

III.              $$-1=100a-10b+c$$

$$-1=-100\cdot \frac{2}{9} +10\cdot 1.7+c$$

$$c=3.\overline{4}$$ !

Sep 22, 2020
edited by asinus  Sep 22, 2020