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## Using Properties of Parabolas to Graph a Parabola Problem 2

As I discussed in an earlier post on graphing a parabola, the following information allows to easily graph a parabola in standard form $f(x)=ax^2+bx+c$.

Properties of a Parabola or a Quadratic Function in Standard Form

Given that $f(x)=ax^2+bx+c$ is a parabola when $a\neq0.$

1. If a > 0, then the graph opens upward. If a < 0, then the graph opens downward.
2. The axis of symmetry is given by the line $-\frac{b}{2a}$.
3. The vertex is given by the ordered pair $(-\frac{b}{2a},f(-\frac{b}{2a}))$
4. Finding 2 other points is easy. Choose any x-value and find the matching y-value. The second point is found by using the axis of symmetry and the location of the 1st points corresponding point.

In my first post the equation did not have a linear term and was in the form $f(x)=ax^2+ c$. The vertex could be found using (0,c) and the axis of symmetry (AOS) of x = 0. In terms of the coordinate plane, the y-axis is the AOS.

In this post, we change gears and will be graphing a parabola that is given in standard form $f(x)=ax^2+bx+c$. The point of this work is to sketch a graph using the core concepts behind the parabolic relation. Remember, because a parabola is symmetric every point except the vertex has a mirror image. Once you find the AOS and vertex, you will have to use the same process to find another point and then its mirror image. Let’s take this to the example.

Example Graphing a Quadratic Equation in Standard Form

Graph $f(x)=2x^2+4x-3$

Looking at the properties above in order, first we know this graph will open upwards, this is due to the fact the coefficient on the squared term is positive or a>0. The AOS is found by using the ratio $-\frac{b}{2a}=-\frac{4}{2(2)}=-1$ and can be written as x = -1. If you are trying to create the graph at the bottom of this post, you should draw a vertical dashed line at x = -1, because this represent the AOS.

Next we find the vertex. Since the equation or function is given as $f(x)=2x^2+4x-3$ and x = -1, it is natural to substitute -1 to get $f(-1)=2(-1)x^2+4(-1)-3$. Using the order of operations to simplify the right side, we arrive at f(-1)=-5. If you have no idea what is going on, here is an explanation. The AOS serves as the x-coordinate of the vertex. Since we know the x-value, we can find the y-value (***f(x) is the same as using the variable y because of ‘function notation’***). By plugging in the known x-value, we can find the related y-value and we now have an ordered pair. For this example the vertex is the ordered pair is (-1,-5). If you are trying to graph this problem, you should plot the point (-1,-5)

Lastly, we need to find two more points. To do this, pick any x-value to the left or right of the AOS. For this example I will use x = -3. Using the given function, $f(x)=2x^2+4x-3$, substituting -3 in for x gives $f(-3)=2(-3)^2+4(-3)-3$ applying the order of operations f(x)=3 and the ordered pair (-3,3). Again, because parabolas are symmetric, (-3,3) has a mirror image at (1,3). If you are graphing along with the post, plot the points and draw a curve to fit those points. Even though this is a sketch, you should try to make it neat.

Graphing a Parabola in Standard Form