Help!? Parabola and its vertex!
1 answer:
Wheeee
ok,
the equation you are given is just a similar form of my vertex form
y=a(x-h)²+k
(we just added k to both sides)
(h,k) is still the vertex
so given that the vertex is (4,-1), h=4 and k=-1
we'll use your way
y+1=a(x-4)²
and the y intercept is (0,15)
plug that point to find the value of a
15+1=a(0-4)²
16=a(-4)²
16=a(16)
1=a
the equation is
expand
y+1=1(x²-8x+16)
y+1=x²-8x+16
y=x²-8x+15
x intercept is where y=0
0=x²-8x+15
factor
0=(x-3)(x-5)
set to zero
0=x-3
3=x
0=x-5
5=x
x intercepts are at (3,0) and (5,0)
ok,
the equation you are given is just a similar form of my vertex form
y=a(x-h)²+k
(we just added k to both sides)
(h,k) is still the vertex
so given that the vertex is (4,-1), h=4 and k=-1
we'll use your way
y+1=a(x-4)²
and the y intercept is (0,15)
plug that point to find the value of a
15+1=a(0-4)²
16=a(-4)²
16=a(16)
1=a
the equation is
expand
y+1=1(x²-8x+16)
y+1=x²-8x+16
y=x²-8x+15
x intercept is where y=0
0=x²-8x+15
factor
0=(x-3)(x-5)
set to zero
0=x-3
3=x
0=x-5
5=x
x intercepts are at (3,0) and (5,0)
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