Graphing quadratics
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Ex 3: Graphing a quadratic function
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Example: Roots and vertex of a parabola
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Example: Parabola vertex and axis of symmetry
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Graphs of Quadratic Functions
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Graphing parabolas in standard form
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Parabola Focus and Directrix 1
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Focus and Directrix of a Parabola 2
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Vertex of a parabola
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Graphing parabolas in vertex form
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Graphing parabolas in all forms
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Parabola intuition 3
Example: Roots and vertex of a parabola Quadratic Functions 3
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- We're asked to graph this function, finding the roots
- and the vertex of this parabola.
- So to find the roots, when people talk about roots,
- they're talking about the points of intersection of this
- function and the x-axis.
- So this is the x-axis here, and this is the y-axis.
- And you're going to intersect the x-axis when
- y is equal to 0.
- So to find the roots, we said y is equal to 0, so we solve
- the equation, x squared plus 4x minus 12 is equal to 0.
- All I did is I said y is equal to 0, and I swapped the left-
- and right-hand sides.
- Now, to factor this, we have a 1 as a leading coefficient, so
- we don't have to do the all out grouping.
- We can just think, are there two numbers whose products are
- negative 12, and if I add them I get 4?
- So let's think about it.
- It looks like 2, positive 6, and negative 2 would work, if
- I go through all of negative 12's factors.
- So this can be factored as x plus 6, times x minus 2.
- 6 minus 2 is 4.
- 6 times negative 2 is negative 12, so that's going to be
- equal to 0.
- And if we have two numbers, and if, when you multiply
- them, they equal 0, that means that either x plus 6 is equal
- to 0, or x minus 2 is equal to 0.
- Subtract 6 from both sides of this equation, you get x is
- equal to negative 6.
- Add 2 to both sides of this equation, and you get x is
- equal to 2.
- So y is equal to 0 when x is equal to negative 6, or x is
- equal to 2.
- So let's graph both of those points.
- x is equal to negative 6, y will be equal to 0.
- So x is 1, 2, 3, 4, 5, 6.
- So negative 6 comma 0.
- That will be on the graph of this function, or the graph of
- this parabola.
- And also the point, x is equal to 2.
- So 1, 2.
- Just like that. x is equal to 2.
- So this is the point 2, 0.
- Now to find the x value of the vertex, there's actually
- several ways of doing it.
- There is a formula that gives it to you.
- But actually, the easiest way to figure it out is it's
- actually going to be the midpoint between the two 0's.
- If you already figured out the 0's, the x value where the
- vertex was going to lie-- because the two 0's are always
- symmetric around the axis of symmetry, which is the same x
- value as the vertex.
- You can just find the midpoint of these two points, and that
- will be the x value of the vertex.
- So let's just do that.
- So if we take the midpoint, we can literally just average
- these two numbers.
- So negative 6 plus 2, over 2 is negative 4 over 2, which is
- equal to negative 2.
- So the vertex is x is equal to negative 2.
- And when x is equal to negative 2, what is y?
- So x is the vertex, and then y will be equal to negative 2
- squared, is positive 4.
- 4 times negative 2 is negative 8.
- And then we have a minus 12 there.
- So 4 minus 8 gets us to negative 4, minus another 12
- is negative 16.
- So the point negative 2, negative 16 is
- going to be our vertex.
- So let me graph that.
- I think it's going to fall off of this graph a little bit.
- So negative 2, and then we're going to go 1, 2, 3, 4, 5, 6,
- 7, 8, 9, 10 11, 12, 13, 14, 15, 16.
- So it's going to be right around here.
- We went off the graph paper.
- But negative 2 all the way down to negative 16.
- So that right there is the vertex.
- Negative 2, negative 16.
- All the way down here.
- We can even lower the y-axis.
- And so if I were going to graph the parabola, it would
- look something like this.
- It would look something like that.
- Just like that and obviously it would keep going.
- Now, I figured out the vertex in this problem just by
- averaging the x values of the two routes.
- That gave me the x value of the axis of symmetry, which is
- this right here. x is equal to negative 2.
- That's the axis around which the parabola is symmetric.
- The other way to figure out the x value of the vertex, is
- to use the formula, negative b over 2a.
- So the x value of the vertex is equal to negative b--
- negative 4-- over 2 times a.
- a is 1, so over 2, which is also going to give you x is
- equal to negative 2.
- Now, the last way to solve for the vertex is to complete the
- square on this parabola right here.
- So what we can do is we can rewrite it.
- We can say this is y is equal to x squared plus 4x.
- I'm going to put the minus 12 out here.
- And what we want to do is add and subtract the same number,
- so I can express this as a perfect square.
- So if I take 1/2 of 4x, or if I take 1/2 of 4, I get 2 if I
- square that.
- I could add a 4 here, and then subtract a 4.
- And then what that does is it turns this into x plus 2
- squared, right? x plus 2 squared is x squared
- plus 4x, plus 4.
- We've seen that multiple times.
- So you have x plus 2 squared.
- And then you have the minus 4 minus 12, minus 16.
- So this is just another way of writing
- this exact same function.
- But when you write it this way you see that the lowest point,
- this part of the function right here is always positive.
- It's a quantity squared.
- The lowest value that it can be is 0.
- But then it can only go up from 0.
- So the lowest point, the lowest y value we can take on,
- is when this expression right here is 0.
- And when that happens, the y value is negative 16.
- Now, what x makes this value 0?
- Well, x equals negative 2 will make this 0.
- If you wanted to say, x plus 2 is equal to 0, this expression
- right here is going to be 0 when x plus 2 is equal to 0.
- Subtract 2 from both sides, x is equal to negative 2.
- So that's another way to find the vertex, all of them
- equally valid.
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