Write the formula for the parabola that has x- intercepts (1.3,0) and (-1.3,0), and y- intercept (0,169).

Step-by-step explanation:

et me start by the following correction:

The y intercept is (0,1.69) and not (0,169)

So the intercept form of the quadratic equation is y=a(x-p)(x-q)

where (-1.3,0)=(p,0) and (1.3,0) is (q,0); then p=-1.3, and q=1.3

With the vertex being at the y axis and the intercepts at the x axis means that the parabola opens down. then 1.69=a(0+1.3)(0-1.3), 1.69=-1.69 a, a=-1

Substituting, y=-1(x+1.3)(x-1.3) note that this is difference of two squares y=-[x2-(1.3)2] , y=-(x2-1.69), y=-x2+1.69

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right