The difficulty of graphing a quadratic function varies depending on the form you find it in. We'll start things off relatively easily.

*y* = *a*(*x* – *h*)^{2} + *k*

No, we're not lying to you; that *is* a quadratic function. Go ahead, multiply it out.

*y* = *ax*^{2} + (-2*ah*)*x* + (*ah*^{2} + *k*)

*h* and *k* are constants, so (-2*ah*) and (*ah*^{2} + *k*) are also constants, which we could call, say, *b *and *c*. See, you can trust us, it's totally quadratic.

When you have a parabola written out in this form, you have several pieces of important information given to you at a glance. When you first meet someone, your first impression tends to stick with you. It's the same with this equation. The sign of *a*, whether it is positive or negative, tells you if the parabola opens up or down. You also know that the vertex of the parabola is at the point (*h*, *k*). Be careful with the sign of *h*, though.

Graph the function *y* = (*x* – 2)^{2} – 1.

The vertex of the parabola is at (2, –1). We can also see that the parabola points upwards. We need a few more points, however. While we could make a table and start plugging in values of *x*, there is usually an easier way: find the intercepts of *y* and *x* (if they exist). Starting with the *y*-intercept, which occurs at *x* = 0.

*f*(0) = (0 – 2)^{2} – 1 = 4 – 1 = 3

(0, 3) is a point on our parabola. Now go for the *x*-intercept(s), which occurs when *y* = 0, if it does so anywhere on this function.

0 = (*x* – 2)^{2} – 1

0 = *x*^{2} – 4*x* + 4 – 1

0 = *x*^{2} – 4*x* + 3

This quadratic equation can be factored.

0 = (*x* – 3)(*x* – 1)

So, (1, 0) and (3, 0) are also points on the parabola. Putting it all together gets us

See? Math smarter, not harder.

Graph the function *y* = -2(*x* + 1)^{2} – 2.

We immediately see that the vertex is at (-1, -2), and the parabola points down. The *y*-intercept is

*f*(0) = -2(0 + 1)^{2} – 2 = -2(1) – 2 = -4

(0, -4). Now we hunt for the *x*-intercepts. Some say the use of dynamite while hunting is unsportsmanlike. They're probably right.

0 = -2(*x* + 1)^{2} – 2

0 = -2(*x*^{2} + 2*x* + 1) – 2

0 = -2(*x*^{2} – 2*x* – 1 – 1)

0 = -*x*^{2} – 2*x* – 2

At this point, we hit a wall. The discriminate of this equation is (-2)^{2} – 4(-1)(-2) = 4 – 8 = -4. There are no real roots for this equation. This means that the function will never cross the *x*-axis, and so there are no *x*-intercepts. This makes sense given that the vertex is at (-1, -2) and the parabola points down, so the function won't go up towards the *x*-axis. Guess we won't need that dynamite after all.

Instead of using the *x*-intercepts, we'll plug in a few extra values of *x* and plot those.

x | y |

3 | -10 |

-4 | -20 |

We're almost ready to finish off this graph. This will be easier to do with a few more points, though. Remember that the axis of symmetry passes through the vertex; we can use this to find several more points now, since we have points on both sides of the vertex.

Compare (0, -4) to the vertex at (-1, -2), for instance. It is 1 to the right on the *x*-axis and 2 lower on the *y*-axis. Because the function is symmetric, one space to the left of the vertex will also be 2 lower on the *y*-axis as well, at (-2, -4). Similarly, (-3, -10) is 2 spaces to the left of the vertex and 8 down, and (-4, -20) is 3 spaces left and 18 spaces down away. So, (1, -10) and (2, -20) are also points on the parabola. Now we can confidently graph the parabola.

When you find yourself graphing a vertex form parabola, here's what you do.

- Check the sign of
*a*to see if it opens up or down.

- Find the vertex and the
*y*-intercept.

- Determine if there are any
*x*-intercepts, by comparing the shape of the parabola and the vertex, or by checking the discriminate of the expanded function.

- Find the
*x*-intercepts if they exist.

- Check if you have enough points to finish the graph. If so, hurray.

- If not, boo. Plot a few more points, and use the function's symmetry across the
*y*-axis to find more points.

- Finish up and call it a day.

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