# Math Help - sketch graph of quadratic equation

1. ## 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

2. 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 !

3. 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.

4. Originally Posted by Moo
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

5. Originally Posted by Jhevon
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

6. Originally Posted by Tweety
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 $x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-5 \pm 7}{4}=\left\{\frac 12,-3\right\}$

7. Originally Posted by Tweety
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})
$

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!