MATH
Lab2-SP12

# Using either substitution or elimination one finds

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Using either substitution or elimination, one finds that a = 4, b = 8 and c = - 12. If you’re observant, you’ll notice that we could save ourselves some work by using the y -intercept to obtain the equation: - 12 = 0 a + 0 b + c = c , which gives us the constant term with much less work. We also have the x -intercepts, but the general form does not give us a way to use them. . . Using general form, we found the equation: y = 4 x 2 + 8 x - 12. 2

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Example 2. Zeros form y = C ( x - z 1 )( x - z 2 ) . If we know the x -intercepts and one other point, then the zeros form is more helpful. We know that our vertical parabola has x -intercepts at x = - 3 and at x = 1. We still need a third point to determine the value of the scaling constant C . Using the x -intercepts, we have: y = C ( x - (- 3 ))( x - 1 ) = C ( x + 3 )( x - 1 ) . Now we use one of the other points – let’s use ( 0, - 12 ) : - 12 = C ( 0 + 3 )( 0 - 1 ) = - 3 C , implying that C = 4. In zeros form, our parabola has the equation: y = 4 ( x + 3 )( x - 1 ) . Example 3. Vertex form y = A ( x - h ) 2 + k . A vertical parabola must have either a highest or a lowest point. This maxi- mum or minimum is called the vertex of the parabola. The parabola is symmetric about the vertical line through the vertex. In our sample vertical parabola, the vertex is at (- 1, - 16 ) . In vertex form, the point ( h , k ) is the location of the vertex. The constant A is a scaling constant. Using the vertex (- 1, - 16 ) , our vertex form becomes: y = A ( x + 1 ) 2 - 16. Now we need just one point to find the scaling constant. For example, using the y -intercept at ( 0, - 12 ) , we see that: - 12 = A ( 0 + 1 ) 2 - 16, implying that A = 4. So, in vertex form, our parabola is given by the equation: y = 4 ( x + 1 ) 2 - 16. Here we only needed two points, but we had the additional information that one of these points was the vertex. Using the symmetry property, our second point ( 0, - 12 ) has a mirror image at (- 2, - 12 ) . Are they equivalent? In this case the equations are equivalent, but if our data had some rounding errors, different forms could lead to inequivalent equations. So how can we determine whether the equations are equivalent? The answer is to write them all in general form using exact arithmetic. 3
Our general form equation was: y = 4 x 2 + 8 x - 12. Our zeros form equation was: y = 4 ( x + 3 )( x - 1 ) . Multiplying the right hand side, we have: y = 4 ( x + 3 )( x - 1 ) = 4 ( x 2 + 2 x - 3 ) = 4 x 2 + 8 x - 12. Our vertex form equation was: y = 4 ( x + 1 ) 2 - 16. Multiplying this out: y = 4 ( x 2 + 2 x + 1 ) - 16 = 4 x 2 + 8 x - 12. All three equations have exactly the same general equation, so they are alge- braically equivalent – as a result, the equations have identical graphs.

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