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    11. Given that f(x) = 2x^2 +8x + 3
    a) find the value of the discriminant.

    Here i worked out the discriminant to be 40.

    b) Express f(x) in the form p(x+q)^2 + r .

    I got 2(x+2)^2 -5 =0

    c) The line y=4x + c is a tangent to the curve y=f(x)

    Calculate the value of C.

    This is where i need help (c) . Thanks.
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    (Original post by siml1)
    11. Given that f(x) = 2x^2 +8x + 3
    a) find the value of the discriminant.

    Here i worked out the discriminant to be 40.

    b) Express f(x) in the form p(x+q)^2 + r .

    I got 2(x+2)^2 -5 =0

    c) The line y=4x + c is a tangent to the curve y=f(x)

    Calculate the value of C.

    This is where i need help (c) . Thanks.
    If you're totally lost, remember that 4x means the gradient at that point is 4. You've got the gradient but you don't have the x and y coordinates - how might you work them out?
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    (Original post by Chlorophile)
    If you're totally lost, remember that 4x means the gradient at that point is 4. You've got the gradient but you don't have the x and y coordinates - how might you work them out?

    Do i have to sub something in somewhere? Thanks for the prompt reply
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    (Original post by siml1)
    Do i have to sub something in somewhere? Thanks for the prompt reply
    You're probably used to differentiating a function to find out the gradient at a certain value of x. However, it's also perfectly possible to make the differentiated function equal to a gradient you're interested in to find out the x-coordinate where the gradient is equal to that. For instance, if I want to know where on the line g(x)=x2 the gradient is equal to 6, I find g'(x)=2x and make that equal to the gradient I'm interested in, i.e. 2x=6. I then solve it, and that tells me at x=3, the gradient is equal to 2. You use this exam same principle here.
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    (Original post by siml1)
    11. Given that f(x) = 2x^2 +8x + 3
    a) find the value of the discriminant.

    Here i worked out the discriminant to be 40.

    b) Express f(x) in the form p(x+q)^2 + r .

    I got 2(x+2)^2 -5 =0

    c) The line y=4x + c is a tangent to the curve y=f(x)

    Calculate the value of C.

    This is where i need help (c) . Thanks.
    a) yes it is 40! 64-4(2)(3)=40

    b) 2x^2 + 8x + 3 = 2(x+2)^2-8+3 = 2(x+2)^2-5 so you're correct!

    c) Well this question links with the discriminant! If we were doing this for a quadratic, we know that if the discriminant = 0 then there is a repeated root, <0 no real roots etc.

    Well we can also use the discriminant for solving the intersections of curves! So this y=4x+c is tangent to f(x), therefore they must intersect!

    So if we let y=y, i.e. 4x + c = f(x), and we will get a quadratic! This tangent will only cut the quadratic (f(x)) in 1 place, but it's repeated root! Therefore we know that b^2-4ac=0, this can be used to find the value of 'c'



    May I ask if this was the C1 from the summer 2014 exams? I remember this question, or one of the same kind aha
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    (Original post by Chlorophile)
    You're probably used to differentiating a function to find out the gradient at a certain value of x. However, it's also perfectly possible to make the differentiated function equal to a gradient you're interested in to find out the x-coordinate where the gradient is equal to that. For instance, if I want to know where on the line g(x)=x2 the gradient is equal to 6, I find g'(x)=2x and make that equal to the gradient I'm interested in, i.e. 2x=6. I then solve it, and that tells me at x=3, the gradient is equal to 2. You use this exam same principle here.
    Right, i kinda get that.
    So i ned to differentiate the f(x), then do 4x+8=4?
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    (Original post by siml1)
    Right, i kinda get that.
    So i ned to differentiate the f(x), then do 4x+8=4?
    Yep! Then what? What pieces of information do you need to work out the equation of a straight line?
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    (Original post by Chlorophile)
    Yep! Then what? What pieces of information do you need to work out the equation of a straight line?
    Right, so x=-1? :confused:
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    (Original post by siml1)
    Right, so x=-1? :confused:
    Yes! What's the next step?
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    (Original post by Chlorophile)
    Yes! What's the next step?
    yeeeey! Erm, i'm not sure...
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    (Original post by siml1)
    Right, so x=-1? :confused:
    There are two ways to do this:

    1) find f'(x), and let that equal the gradient of the y=4x+c, i.e. =4. Then whatever value you get, find the corresponding y value from f(x) and then sub in this coordinate where the tangent is into y=4x+c to find the value of c!

    OR

    2) like when finding the intersection of 2 lines, let y=y i.e. f(x) = 4x + c, which results in a quadratic. Because it's tangent, that implies a repeated root of their intersections (1 root), therefore do b^2-4ac=0 and solve for c!
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    (Original post by siml1)
    yeeeey! Erm, i'm not sure...
    Okay, well there are several ways you could continue this but the easiest way is probably this. I'm assuming you know how to work out the equation of a straight line? We have the gradient and an x coordinate, what one other piece of data do we need to work out the equation of the straight line?

    ^Chazatthekeys has answered the question for you! Method 1 is probably easier but both are fine.
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    (Original post by Chazatthekeys)
    There are two ways to do this:

    1) find f'(x), and let that equal the gradient of the y=4x+c, i.e. =4. Then whatever value you get, find the corresponding y value from f(x) and then sub in this coordinate where the tangent is into y=4x+c to find the value of c!

    OR

    2) like when finding the intersection of 2 lines, let y=y i.e. f(x) = 4x + c, which results in a quadratic. Because it's tangent, that implies a repeated root of their intersections (1 root), therefore do b^2-4ac=0 and solve for c!


    so...c=8?!
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    (Original post by siml1)
    so...c=8?!
    Not quite! If we let f'(x) = 4 (because the gradient of y=4x+c is 4) we get 4x+8 = 4, therefore x=-1 like you stated!

    We know need to find the corresponding y-coord for this point on the curve and tangent!

    y=f(-1)=2(-1)^2+8(-1)+3=2-8+3=-3

    Now we know that the tangent is a tangent to f(x) at (-1,-3) which we can sub into y=4x+c to find the value of c!
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    (Original post by Chazatthekeys)
    Not quite! If we let f'(x) = 4 (because the gradient of y=4x+c is 4) we get 4x+8 = 4, therefore x=-1 like you stated!

    We know need to find the corresponding y-coord for this point on the curve and tangent!

    y=f(-1)=2(-1)^2+8(-1)+3=2-8+3=-3

    Now we know that the tangent is a tangent to f(x) at (-1,-3) which we can sub into y=4x+c to find the value of c!
    Oh right, so c=1!
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    (Original post by siml1)
    Oh right, so c=1!
    Yes! Wahoo! lol
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    (Original post by Chazatthekeys)
    Yes! Wahoo! lol

    Thankyou for the help. (both of you).

    Have a nice day.

    Siml1.
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    (Original post by siml1)
    Thankyou for the help. (both of you).

    Have a nice day.

    Siml1.
    Don't mention it you too!
 
 
 
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