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C1 help!

Given that f(x) = x^2-6x+18, x >= 0 (greater than = to)

a). express f(x) in the form (x-a)^2+b where a and b are integers.
This one I could do, it was just completing the square so I ended up with
(x-3)^2+9. So a is 3 and b is 9.

b). The curve c with equation y = f(x) x >= 0 meets the y-axis at P and has a minimum point at Q.
Sketch the graph of C, showing the coordinates of P and Q. ( can someone explain how to do this)

The line y= 41 meets C at the point R

C). Find the x-coordinate of R giving your answer in the form p+q root 2, where p and q are integers. ( lost on this as well).

2). Given that the equation kx^2 + 12x + k = 0, where k is a positive constant, has equal roots, find the value of k. I missed the lesson on this so if someone can explain how to do this question it would be much appreciated. Thanks :smile:
(edited 10 years ago)
Reply 1
Avatar for joe1545
joe1545
14
Original post by Super199
Given that f(x) = x^2-6x+18, x >= 0 (greater than = to)

a). express f(x) in the form (x-a)^2+b where a and b are integers.
This one I could do, it was just completing the square so I ended up with
(x-3)^2+9. So a is 3 and b is 9.

b). The curve c with equation y = f(x) x >= 0 meets the y-axis at P and has a minimum point at Q.
Sketch the graph of C, showing the coordinates of P and Q. ( can someone explain how to do this)

The line y= 41 meets C at the point R

C). Find the x-coordinate of R giving your answer in the form p+q root 2, where p and q are integers. ( lost on this as well).

2). Given that the equation kx^2 + 12x + k = 0, where k is a positive constant, has equal roots, find the value of k. I missed the lesson on this so if someone can explain how to do this question it would be much appreciated. Thanks :smile:


For B)

Are you fammiliar with the transformation of graphs, so if i said to you f(x) = x2
would you know what f(x-3) looks like.
Basically the x-3 bit shifts the graph 3 places to the right. And if you have f(x) + 9 you would shift the curve 9 places up the y axis.

So essentially (x-3)^2 + 9 = f(x-3) + 9 to find where it cuts the Y axis put in x = 0
and the minimum point, which is the lowest point on the curve can be seen if you sketch the graph.

I hope thats clear, if not use a graph plotter online.

As for C when lines intersect each other the two lines equel each other at this particular point.. So y=y hence 41= (x-3)^2 +9, if you open out all the brackets to get it as a quadratic and make it = 0 you should be able to solve it. As it wants the answer as a surd this indicates you will probably have to use the quadratic formula for this.
Reply 2
Avatar for joe1545
joe1545
14
Original post by Super199
Given that f(x) = x^2-6x+18, x >= 0 (greater than = to)

a). express f(x) in the form (x-a)^2+b where a and b are integers.
This one I could do, it was just completing the square so I ended up with
(x-3)^2+9. So a is 3 and b is 9.

b). The curve c with equation y = f(x) x >= 0 meets the y-axis at P and has a minimum point at Q.
Sketch the graph of C, showing the coordinates of P and Q. ( can someone explain how to do this)

The line y= 41 meets C at the point R

C). Find the x-coordinate of R giving your answer in the form p+q root 2, where p and q are integers. ( lost on this as well).

2). Given that the equation kx^2 + 12x + k = 0, where k is a positive constant, has equal roots, find the value of k. I missed the lesson on this so if someone can explain how to do this question it would be much appreciated. Thanks :smile:


For question 2 you have to use the discrimminant, which is b2-4ac
you should know that for equel roots b2 = 4ac

Hence 122 = 4(k)(k) or 144 = 4K2

Now you should be able to solve that equation to find K on its own
B) The minimum value of (x-3)^2 + 9 is 9. Why is this? At what x does it take that value?
Reply 4
Avatar for Super199
Super199
OP
20
Original post by joe1545
For question 2 you have to use the discrimminant, which is b2-4ac
you should know that for equel roots b2 = 4ac

Hence 122 = 4(k)(k) or 144 = 4K2

Now you should be able to solve that equation to find K on its own

I understand! But one thing I don't get is that I got k=6 but the book says the x = -1? How did they know it was -1?
Original post by Super199
Given that f(x) = x^2-6x+18, x >= 0 (greater than = to)

a). express f(x) in the form (x-a)^2+b where a and b are integers.
This one I could do, it was just completing the square so I ended up with
(x-3)^2+9. So a is 3 and b is 9.

b). The curve c with equation y = f(x) x >= 0 meets the y-axis at P and has a minimum point at Q.
Sketch the graph of C, showing the coordinates of P and Q. ( can someone explain how to do this)

The line y= 41 meets C at the point R

C). Find the x-coordinate of R giving your answer in the form p+q root 2, where p and q are integers. ( lost on this as well).

2). Given that the equation kx^2 + 12x + k = 0, where k is a positive constant, has equal roots, find the value of k. I missed the lesson on this so if someone can explain how to do this question it would be much appreciated. Thanks :smile:


When completing the square you have forgotten about the +18.
Reply 6
Avatar for Super199
Super199
OP
20
Original post by brianeverit
When completing the square you have forgotten about the +18.

Have I? (x-3)^2 -3^2 which is -9 + 18 = 9?
Original post by Super199
Have I? (x-3)^2 -3^2 which is -9 + 18 = 9?


Sorry .
Original post by electriic_ink
B) The minimum value of (x-3)^2 + 9 is 9. Why is this? At what x does it take that value?


MInimum possible value of (x-3)^2 is 0 when x=3
Reply 9
Avatar for Super199
Super199
OP
20
Original post by brianeverit
MInimum possible value of (x-3)^2 is 0 when x=3


How would i do this one? X^2-6x+a =(x+b)^2
Find the value of a and the value of b

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