I've tried to complete this question for like an hour and even asked other people . They got similar answers to me.
the questions are finding the vertex from an equation
the first one is -x^2 - 7x -6
this one is totally off the vertex point of -6
I got something like ( 7/2 , -6.25) for the vertex
for # 2 y= 2x^2 -5x - 7
the vertex I got was ( 5/4 , -81/8 ) with the stretch being 1/2 . this was off by one point
so please help because this is an assignment I have to pass it in and it is worth quite a bit of marks and the answers are not working out even though I followed the steps of how to do it. basically this question is first factoring ( 2 step composition for second one and first you take - out of equation) to get the x intercepts then you have to complete the square to get the vertex. if you could please answer this question I'd appreicate it so much! especially if you show your work. Graphing your vertex after would be helpful as it needs to pass through the y intercept. thanks to those who answer it :).
the questions are finding the vertex from an equation
the first one is -x^2 - 7x -6
this one is totally off the vertex point of -6
I got something like ( 7/2 , -6.25) for the vertex
for # 2 y= 2x^2 -5x - 7
the vertex I got was ( 5/4 , -81/8 ) with the stretch being 1/2 . this was off by one point
so please help because this is an assignment I have to pass it in and it is worth quite a bit of marks and the answers are not working out even though I followed the steps of how to do it. basically this question is first factoring ( 2 step composition for second one and first you take - out of equation) to get the x intercepts then you have to complete the square to get the vertex. if you could please answer this question I'd appreicate it so much! especially if you show your work. Graphing your vertex after would be helpful as it needs to pass through the y intercept. thanks to those who answer it :).
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(y-k)=s(x-h)^2 in this form we find the vertex (h,k) and how stretched it is s
now to get these into this form we do a few steps 1 get x^2 term to equal 1
so
1)
y=-x^2-7x-6
-y=x^2+7x+6
2)
y=2x^2-5x-7
y/2=x^2-2.5-3.5
next get rid of the end coefficient
1)
-y-6=x^2+7x
2)
.5y+3.5=x^2-2.5x
next complete square on right by adding the square of half the x term to both sides
1)
-y-6+12.25=x^2+7x+12.25
-y+6.25=[x+(7/2)]^2
2)
(1/2)y+(7/2)+(5/4)^2=x^2-2.5x+(5/4)^2
(1/2)y+81/16=[x-(5/4)]^2
next set y coefficient equal to 1
1)
(y-6.25)=-[x+(7/2)]^2
2)
y+81/8=2[x-(5/4)]^2
now we got in form we wanted just pull out h and j
1)
(-7/2,6.25)
2)
(5/4,-81/8)
now to get these into this form we do a few steps 1 get x^2 term to equal 1
so
1)
y=-x^2-7x-6
-y=x^2+7x+6
2)
y=2x^2-5x-7
y/2=x^2-2.5-3.5
next get rid of the end coefficient
1)
-y-6=x^2+7x
2)
.5y+3.5=x^2-2.5x
next complete square on right by adding the square of half the x term to both sides
1)
-y-6+12.25=x^2+7x+12.25
-y+6.25=[x+(7/2)]^2
2)
(1/2)y+(7/2)+(5/4)^2=x^2-2.5x+(5/4)^2
(1/2)y+81/16=[x-(5/4)]^2
next set y coefficient equal to 1
1)
(y-6.25)=-[x+(7/2)]^2
2)
y+81/8=2[x-(5/4)]^2
now we got in form we wanted just pull out h and j
1)
(-7/2,6.25)
2)
(5/4,-81/8)
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Take the derivative of the polynomial, first, then set equal to zero:
-2x - 7 = 0
Solve: -2x = 7
x = -3.5
Plug -3.5 in for x in the original polynomial, and evaluate to get the y coordinate:
- (-3.5)^2 -7 (-3.5) -6
-12.25 + 24.5 - 6 = 6.25
Vertex is (-3.5, 6.25)
-2x - 7 = 0
Solve: -2x = 7
x = -3.5
Plug -3.5 in for x in the original polynomial, and evaluate to get the y coordinate:
- (-3.5)^2 -7 (-3.5) -6
-12.25 + 24.5 - 6 = 6.25
Vertex is (-3.5, 6.25)
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Why don't you go to school tomorrow and ask the teacher to explain it to you.