Unit
Quadratic Equations and Functions
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
One thing you guys probably already know about graphs is that they’re useful in lots of ways, but sometimes they’re not very useful. Like if you had a vertex that was a fractional value, it’d be really hard to look at a graph and be able to tell what fraction was being represented. That’s why it’s really important that you know how to find the vertex algebraically as well as how to find it graphically.
So let’s go with this problem to find the vertex, I’m going to start by figuring out what my x coordinate is, -b/2a. In our case I’ll have -3 over 2 times 1 which is -3/2, that’s going to be the x coordinate of my vertex. To find the corresponding y value, I’m going to have to substitute that fraction back in. Y is going to be -3/2 squared plus 3 times -3/2 plus 4. I can already tell this is common denominator coming on. -3/2 times itself is 9/4, 3 times 3/2 is -9/2 there and then plus 4.
In order to do that addition, I need to find the common denominator, so this first fraction I’m going to leave is 9/4, the second guy I’m going to turn into -18/4, and then this I’m going to turn into 16/4, and that was good because now I can just look at the numerator add across the top from left to right. 9 take away 18 is -9 plus 16 is+7, so my y value is going to be 7/4. This is my vertex right here.
I could verify that by looking at a graph to just make sure that it’s somewhere in between the x values -1 and -2 and somewhere in between the y values, of +1 and +2.
The next thing this problem asked me to do is to write the equation for the Axis of Symmetry, but don’t worry it’s not hard we’ve pretty much already done it. Remember the equation for the Axis of Symmetry looks like x equals -b/2a. It’s a line and -b/2a is how I found that coordinate. So my Axis of Symmetry is the line x equals -3/2, that’s a vertical line that goes right up and down through my parabola. The Axis of Symmetry is really helpful when you’re graphing points.