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Math Help - sketch graph of quadratic equation

  1. #1
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    sketch graph of quadratic equation

    sketch the graph of the following equation

     y= 6-10x-4x^2


    I rearranged the equation to y=  -4x^2 -10x +6

    but i am stuck here cause can't seem to factor the equation, and i am not meant to use the quadratic equation or completing the square.

    how do i facto this?

    i tried  (2x-3) ( -2x-2) =0 but that does not work
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  2. #2
    Moo
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    Hello,

    So can you use the discriminant ?

    \Delta=100+4 \times 4 \times 6=14^2
    And it factorises to 2(x+3)(2x-1)

    So you have the two x-intercepts.
    Now you need the vertex (where the derivative is 0) to draw the graph !
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  3. #3
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    You should find the maxima/minima and intersections with x-axis. You can rewrite the function:
    y=6-10x-4x^2=-2\left(2x^2+5x+\frac{25}{8}-\frac{25}{8}-3\right)=-2\left[\left(x\sqrt{2}+\frac{5}{2\sqrt{2}}\right)^2-\frac{49}{8}\right]
    And the final :-)
    y-\frac{49}{4}=-2\cdot\left(x\sqrt{2}+\frac{5}{2\sqrt{2}}\right)^2
    So there is minus on the right side of the equation so it is parabola inverted by x-axis. Its maximum is in point
    X_{\max}=\left[-\frac{5}{4};\frac{49}{4}\right]

    Intersection points you'll find by discriminant.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    So can you use the discriminant ?

    \Delta=100+4 \times 4 \times 6=14^2
    And it factorises to 2(x+3)(2x-1)

    So you have the two x-intercepts.
    Now you need the vertex (where the derivative is 0) to draw the graph !
    Since this is in the pre-algebra/algebra section, lets assume the OP doesn't know how to find derivatives (i hardly think s/he would have problems graphing parabolas were that the case).

    for a quadratic of the form y = ax^2 + bx + c the vertex occurs where x = \frac {-b}{2a}. of course you can find the corresponding y-value by plugging in the x-value for the vertex
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    Since this is in the pre-algebra/algebra section, lets assume the OP doesn't know how to find derivatives (i hardly think s/he would have problems graphing parabolas were that the case).

    for a quadratic of the form y = ax^2 + bx + c the vertex occurs where x = \frac {-b}{2a}. of course you can find the corresponding y-value by plugging in the x-value for the vertex
    I am not sure what everyone here is suggesting, I have not been taught all that stuff yet.

    basically I am meant to sketch the graph by firstly putting y=0 to find the x-axis crossing points coordinates and than put x=0 to find the y crossing points coordinates.

    and cause  b^2>4ac and  a<0 there are two different roots.

    but i can't daw the graph as i am not able to factorise the eqution to solve for x ?

    does anybody understand what i am meant to do now?

    thanks
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  6. #6
    Moo
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    Quote Originally Posted by Tweety View Post
    I am not sure what everyone here is suggesting, I have not been taught all that stuff yet.

    basically I am meant to sketch the graph by firstly putting y=0 to find the x-axis crossing points coordinates and than put x=0 to find the y crossing points coordinates.

    and cause  b^2>4ac and  a<0 there are two different roots.

    but i can't daw the graph as i am not able to factorise the eqution to solve for x ?

    does anybody understand what i am meant to do now?

    thanks
    b^2-4ac is called the discriminant.

    -4x^2-10x+6=0 \implies 2x^2+5x-3=0
    b^2-4ac=5^2+4 \times 3 \times 2=49=7^2

    And we know that the roots are x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-5 \pm 7}{4}=\left\{\frac 12,-3\right\}
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  7. #7
    Member TheMasterMind's Avatar
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    Quote Originally Posted by Tweety View Post
    sketch the graph of the following equation

     y= 6-10x-4x^2


    I rearranged the equation to y=  -4x^2 -10x +6

    but i am stuck here cause can't seem to factor the equation, and i am not meant to use the quadratic equation or completing the square.

    how do i facto this?

    i tried  (2x-3) ( -2x-2) =0 but that does not work
    The first thing i would do is complete the square to see where the vertex is.
    (I'm sure you know how to do that)

    I get

    y= -4(x+\frac{5}{4})^2 + \frac{49}{4}

    V=(-\frac{5}{4},\frac{49}{4})<br />

    Next I would make a table of values

    Ex take the x values -4,-3,-2,-1,0,1,2 and plug it into the equation to solve for the different y values. (there will not be the exact vertex co-ordinates, although it will be very close.) Then simply plot the points and you're done!
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