Quadratics

Question

Given y= -2x²- 12x what formula would i need to use in order to graph that. 

Answer 1

The graph of y = Ax²+Bx+C is a parabola.  C is the y-intercept.             If A > 0 , the graph opens upward and if A < 0 , the parabola opens downward.  The highest or lowest point of the parabola is called the vertex.  The x-coordinate of the vertex is -B/(2A).

 

So, for y = -2x² - 12x  A = -2, B = -12, C = 0

 

y-intercept = C = 0

A < 0, so the graph opens downward and has a highest point

x-coordinate of vertex (highest point) = -(-12)/(2(-2)) = -3

y-coordinate of vertex = -2(-3)²-12(-3) = 18

 

The graph is a parabola opening downward with highest point (-3, 18).

The graph is symmetric with respect to the vertical line x = -3.  So, by symmetry, since (0,0) is on the graph, so is (-6,0).

 

Now, plot the points (-3,18), (0,0) and (-6,0) and sketch the parabola.  

 

 

Answer 2

You already have the formula!

 

Create a table of points for x, and solve for y at each point.

 

 

x                y

-2    -2(-2)²- 12(-2) = 16

-1    -2(-1)²- 12(-1) = 10

0     -2(0)²- 12(0) = 0

1     -2(1)²- 12(1) = -14

2     -2(2)²- 12(2) = -32

 

Then plot each (x,y) pair.

 

(-2,16)

(-1,10)

(0,0)

(1,-14)

(2,-32)