1) write an equation for the parabola whose
vertex is at (-8,4) and passes through (-6,-2)
The equation of a parabola with vertex (h,k) is
y = a(x - h)² + k
Substitute in (h,k) = (-8,4)
y = a(x - (-8) )² + (4)
y = a(x + 8)² + 4
Now we can substitute (x,y) = (-6,-2)
(-2) = a(-6 + 8)² + 4
-2 = a(2)² + 4
-2 = a(4) + 4
-2 = 4a + 4
-6 = 4a
= a
= a
So substitute for a in
y = a(x + 8)² + 4
and we have
y = (x + 8)² + 4
Drawing the graph:
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2) i need to write: y = x² + 4x - 1 in vertex form
y = x² + 4x - 1
factor 1 out of the first two terms on the right:
y = 1(x² + 4x) - 1
Complete the square by
1. taking 1/2 of the coefficient of x
2. squaring that quantity
3. Adding it and subtracting it inside the
parentheses.
1. 1/2 of 4 is 2
2. 2 squared is 4
3.
y = 1(x² + 4x + 4 - 4) - 1
Change the parentheses to brackets:
y = 1[x² + 4x + 4 - 4] - 1
Factor the frirst three terms inside the
bracket:
y = 1[(x + 2)(x + 2) - 4] - 1
Write (x + 2)(x + 2) as (x + 2)²
y = 1[(x + 2)² - 4] - 1
Remove the bracket by distributing the 1
into the bracket, leaving the parentheses
intact.
y = 1(x + 2)² - 4 - 1
Combine the last two terms
y = 1(x + 2)² - 5
Copare that to
y = a(x - h)² + k
and you can see that a = 1, h = -2, k = -5
so the vertex is (h,k) = (-2,-5)
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3) which quadratic function has its vertex at (-2,7)
and opens down?
Well, many many poarabolas have that vertex and
open down. We'll find one.
Substitute (h,k) = (-2,7) in the standard equation
y = a(x - h)² + k
y = a(x - (-2) )² + 7
y = a(x + 2)² + 7
But then choose any negative number for a, and its
graph will open downward. For instance, letting
a = -1 gives this parabola:
y = -(x + 2)² + 7
Edwin