TABLE OF CONTENTS
Graph the equation y = x2 + 2.
Here are some points:
Here is a graph:
Connecting the dots in a "U'' shape gives us
This looks almost exactly like the graph of y = x2, except we've moved the whole picture up by 2. We like the way it looks up there better. Brings the whole graph together, wouldn't you say?
Graph the quadratic equation y = -x2 + 2x + 3.
We know what we have to find, so let's find it.
1. What are the intercepts?
First, the x-intercepts. We need to find the roots of the quadratic polynomial. If we find them, we can celebrate by drinking a root beer.
We need to find the solutions to the equation
0 = -x2 + 2x + 3 = -(x2 – 2x – 3).
This equation factors as
0 = -(x – 3)(x + 1),
so the solutions (and the x-intercepts) are
x = 3, x = -1.
We can graph these points:
The y-intercept is the constant term, 3, so we can graph that also:
2. What is the vertex?
The vertex occurs halfway between the x-intercepts -1 and 3, so at x = 1. When we plug x = 1 in to the quadratic equation, we find
-(1)2 + 2(1) + 3 = 4,
so the vertex occurs at (1, 4).
3. Does the parabola open upwards or downwards?
Since the coefficient on the x2 term is negative, the parabola opens downwards.
Putting together all the pieces, we find our graph:
We know this graphing stuff can be infectious, but be careful. We don't want you to get a graph infection.
Graph the quadratic equation y = 3x2 + x – 3.
Let's find the useful things. Other than duct tape, a Swiss army knife and a compass. We meant more within the realm of algebra. We should have been clearer.
To find the x-intercepts, we need to use the quadratic equation because this polynomial doesn't factor nicely. We find
For the purposes of graphing, we can round these numbers to 0.8 and -1.2:
The y-intercept is the constant term of the quadratic equation, or -3:
The vertex is at , which in this case is .
When we plug back into the equation, we find
The vertex is .
For the sake of graphing, as well as for the sake of our own sanity, we can round to (-0.2, -3.1).
Since the coefficient on the x2 term is positive, the parabola opens upwards. Like a blossoming flower. That's so nice.
Putting everything together, we get our graph: