suppose a parabola has an axis of symmetry at x = -8, a maximum height of 2, and passes through the point (-7, -1). Write the equation of the parabola in vertex form

Step-by-step explanation:

f(x) = a(x - h)² + k   - the vertex form of the quadratic function with vertex    (h, k)

the axis of symmetry at x = -8 means h = -8

the maximum height of 2 means  k = 2

So:

f(x) = a(x - (-8))² + 2

f(x) = a(x + 8)² + 2   - the vertex form of the quadratic function with vertex   (-8, 2)

The parabola passing through the point (-7, -1) means that if x = -7 then        f(x) = -1

so:

    -1 = a(-7 + 8)² + 2

 -1 -2 = a(1)² + 2 -2

      -3 = a

Threfore:

The vertex form of the parabola which has an axis of symmetry at x = -8, a maximum height of 2, and passes through the point (-7, -1) is:

                                 f(x) = -3(x + 8)² + 2

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!