# How to sketch the parabola "y=-2x^2-4x+6" without making a table of values ?

What you need to do is put this equation that is currently in "general form" (y=ax^2 +bx+c) into what is called "vertex form" (y-k=a(x-h)^2)

To do this, you must follow these steps:

1)  Subtract "c" from both sides of the equation.

2)  Factor the quadratic coefficient ("a") from the right...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

What you need to do is put this equation that is currently in "general form" (y=ax^2 +bx+c) into what is called "vertex form" (y-k=a(x-h)^2)

To do this, you must follow these steps:

1)  Subtract "c" from both sides of the equation.

2)  Factor the quadratic coefficient ("a") from the right side of the equation.

3)  Complete the square of the binomial in parentheses on the right( divide the linear term by 2 and square it; make sure you multiply it by the the factor from part 2 (If necessary and add it to the left)

4)  Factor the resulting "perfect square trinomial" into a "binomial squared"

What you should now have is an equation of the form "y-k=a(x-h)^2.  The point (h, k) will be the vertex of your functions.  The sign of "a" will determine the direction of your  "parabola" To graph the width; pick a value of x that is 1 higher than the x coordinate of the vertex and then "plug it into to get the "y" value.  Then pick an x-value that is one lower and do the same thing.  Now you have enough info to draw a rough estimate of the "parabola"

1) subtracted 6 from both sides.

y-6=-2x^2-4x

2)  I then factored out -2 from the left hand side.

y-6=-2(x^2+2x         )

*(Noticed how I left some space in my parentheses to "complete the square")

3)  I completed the square on the right and added 2 to the right (-2X1)

y-6-2=-2(x-2x+1)

y-8=-2(x+1)^2

The vertex of your function is (-1, 8) and contains the points (0, 6) and (-2, 6).

Approved by eNotes Editorial Team

y= -2x^2 - 4x + 6

First we take a look at the parabola, we notice that the coefficient of x^2 is negative.

Then, we know that the parabola is facing down.

Then, The parabola has a maximum value.

We will calculate the maximum values.

We will find the first derivative y'

==> y' = -4x - 4 = 0

==> x = -1

==> y(-1) = -2 +4 + 6 = 8

Then, the maximum value is at the point ( -1, 8)

Now we will determine the intersection points between the curve and the x.axis.

==> y= -2x^2 -4x + 6 = 0

==> we will divide by -2.

==> x^2 + 2x - 3 = 0

==> (x+3)(x-1) = 0

==> x1= -3

==> x2= 1

Then, the curve  y intersect with the x-axis at the points ( -3,0) and (1,0)

Now we can sketch the parabola using the following points:

Maximum point ( -1, 8)

Intersection with x-axis: ( 1,0) and (-3, 0)

Approved by eNotes Editorial Team