f(x)=2x^2+0x-5
Find the vertex
0/4=0=x
Input x into the equation to find y or f(x)
0+0-5=f(x)
(0,-5) = Vertex
Now find a random x coordinate to input into the equation.
Let's do 4.
32+0-5=f(x)=27
It may not look like a parabola due to it being uneven.
Answer:
The absolute value can be graphed using the points around the vertex (-2, 0), (-4, 2), (-3, 1), (-1, 1), (0, 2).
Step-by-step explanation:
The find the x coordinate of the vertex, set the inside of the absolute value x + 2 equal to 0. In this case, x + 2 = 0.
x + 2 = 0
Subtract 2 from both sides of the equation.
x = -2
Replace the variable x with -2 in the expression.
y = |(-2) + 2|
The absolute value vertex is (-2, 0).
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
For each x value, there is on y value. Select few z values from the domain. It would be more useful to select the values so that they are around the x value of the absolute value vertex.
Substitute the x value -4 into f(x) = |x+2|. In this case, the point is (-4,2).
y=2
Substitute the x value -3 into f(x) = |x+2|. In this case, the point is (-3,1).
y=1
Substitute the x value 0 into f(x) = |x+2|. In this case, the point is (0,2).
y=2
You can find the graph in the attachment.
