University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Finding some more information about a quadratic equation is basically sort of a set formula approach. For finding our intercepts, we just plug in the opposite variable as 0 and then we also have some other things for our vertex, domain and range.
So let's start with our x-intercept. First thing we do when we're finding an x-intercept is let y equals 0, so we plug y is equal to 0 and we end up with 0 is equal to x² plus 3x minus 10 and all we have to do is solve this out. Often times we'll be able to factor this which will make our life easy, sometimes we'll have to use the quadratic formula which not so easy, but again we know how to do. This example factors giving us x plus 5 and x minus 2, so we're able to find our x intercepts are -5 and 2.
Going on to our y intercepts. Whenever we're finding our y intercepts, we just let x equals 0. In this case is going to be pretty easy because what we find out is our first two terms cancel out just leaving us with -10. Our vertex, for our vertex we have options. We could either complete the square which is going to take a couple more steps and it will be a little more work, or we can just use -b over 20. -b over 2a in this case will give us -3 over 2 which is going to be the x coordinate of our vertex. To find the y coordinate, we just have to plug this in and see what comes up, so we plug in -1 and a half, we end up with 1/2 plus 3 times -1 point 5, minus 10 is equal to -12 and a quarter.
So we were able to find our vertex. Our domain for our parabola, domain is what x values can be, we don't have any restrictions on what x can be, so we simply have to say all reals and our range is the y values. This is an upward facing parabola, so we know that the vertex is going to be the lowest point, range refers to y value, so all we're going to be dealing with is -12 and a quarter and up and -12 and a quarter is on the curve, so we include it with a hard bracket.
So finding some information about a parabola, intercepts just solving equations out, vertex -b over 2a plug it in to find the y coordinate, domain is all reals, range just take into consideration your y coordinate of the vertex and if the graph is facing up or down.
Unit
Quadratic Equations and Inequalities