Question

4. Determine the equation of the parabola with x-intercepts

a) -4 and 3, and that passes through (2, 7)

b) 0 and 8, and that passes through (-3, -6)

c) √7 and - √7, and that passes through (-5, 3)


5. Determine the equation of the parabola with vertex

a) (-2, 5) and that passes through (4, -8)

Answer 1

4.

a.) -7/6 (x + 4)(x - 3) = y

b.) -6/121 (x-8)^2 = y

c.) 1/6 (x-sqrt 7)(x + sqrt 7) = y

5.

a.) -13/36 (x+2)^2 + 5 = y

Explanation:

4. The intercept form of a quadratic equation is

`y=a(x-p)(x-q)`, where a is a number determining things like if the parabola opens up or down and how wide or narrow the parabola, while p and q are the x-intercepts.

Because we're given the x-intercepts and a point on the parabola, we can simply plug everything in and solve for a to find the equation of the parabola:

a.):

`y=a(x-(-4))(x-3)\\y=a(x+4)(x-3)\\\\7=a(2+4)(2-3)\\7=a(6)(-1)\\7=-6a\\-7/6=a\\\\y=-7/6(x+4)(x-3)`

b.):

`y=a(x-8)^2\\\\-6=a(-3-8)^2\\-6=a(-11)^2\\-6=121a\\-6/121=a\\\\y=-6/121(x-8)^2`

c.):

`y=a(x-√(7))(x-(-√(7)))\\ y=a(x-√(7))(x+√(7))\\ \\ 3=a(-5-√(7))(-5+√(7))\\ 3=a(25-5√(7)+5√(7)-7)\\ 3=a(25-7)\\ 3=18a\\ 3/18=1/6=a\\ \\ 1/6(x-√(7))(x+√(7))=y`

5. The vertex form of a quadratic equation is

`y=a(x-h)^2+k`, where a is the same type of number as in the intercept form and (h, k) is the vertex.

Like the intercept form of the equation, we're given everything except a and thus we can plug in what we have and solve for a:

`y=a(x-(-2))^2+5\\y=a(x+2)^2+5\\\\-8=a(4+2)^2+5\\-8=a(6)^2+5\\-8=36a+5\\-13=36a\\-13/36=a\\\\y=-13/36(x+2)^2+5`