Transformations of a parabola.

• Nov 10th 2011, 03:19 PM
squelia
Transformations of a parabola.
So here is the question.

y = (3 -x)(x +1) is an inverted parabola. Coordinates are (-1,0) and (3,0) for the x-intercepts, and the y intercept is (0,3). Axis of symmetry is (1,0).

The parabola is moved 2 units to the left and 7 units down. Give the equation for the parabola in its new position and give the coordinates of the y-intercept.

If you can answer this, explain it with as simple terms as possible, please. (Itwasntme)
• Nov 10th 2011, 03:32 PM
skeeter
Quote:

Originally Posted by squelia
So here is the question.

y = (3 -x)(x +1) is an inverted parabola. Coordinates are (-1,0) and (3,0) for the x-intercepts, and the y intercept is (0,3). Axis of symmetry is (1,0).

The parabola is moved 2 units to the left and 7 units down. Give the equation for the parabola in its new position and give the coordinates of the y-intercept.

If you can answer this, explain it with as simple terms as possible, please. (Itwasntme)

\$\displaystyle y = -x^2 + 2x + 3\$

\$\displaystyle y = -(x^2 - 2x) + 3\$

complete the square to get the quadratic in vertex form, \$\displaystyle y = a(x-h)^2 + k\$ ...

\$\displaystyle y = -(x^2 - 2x + 1) + 3 + 1\$

\$\displaystyle y = -(x -1)^2 + 4\$

translate 2 units left, 7 units down ...

\$\displaystyle y = -[(x-1) + 2]^2 + 4 - 7\$

\$\displaystyle y = -(x + 1)^2 - 3\$

evaluate at \$\displaystyle x = 0\$ to find the y-intercept

btw ... wrong forum.