Quadratic Formulas, Equations, Parabolas, GraphingDate: 6/16/96 at 22:30:33 From: Glen McKenzie Subject: Quadratic Problem My name is Brooke McKenzie. I'm in grade ten math and I was wondering if you could help me with a couple problems. We are doing Quadratic Formulas, Equations, and Parabolas, and Graphing them. The first equation is: 6x(squared) + x - 1 = 0 (I'm not sure how to say that just the x is squared) *I need to find the roots (x-intercepts), the vertex, some coordinates and graph it.* I am mostly having problems graphing because I end up with numbers like 1 1/24, -1/12, 14 and so on. I'll show you how I worked it out and maybe you can see if I'm doing something wrong or whatever. FINDING ROOTS 6x^2 + x = 0 (2x + 1)(3x - 1) 2x+1=0 3x-1=0 x= -1/2 x= 1/3 roots (-1/2, 0) (1/3, 0) FINDING VERTEX m= -1/2 + 1/3 y=6 (x squared) + x - 1 --------- =6(-1/12 squared) + (-1/12) -1 2 =6 (1/144) - 1/12 - 1 = 3/6 + 2/6 =6/144 - 1/12 - 1 --------- =1/24 - 2/24 - 24/24 2 =-1/24 - 24/24 = -1/6 =-25/24 ------ = -1 1/24 2 =-1/6 x 1/2 vertex (-1/12, -1 1/24) =-1/12 FINDING COORDINATES these are what I used: x y 1 6 3/2 14 0 -1 -1 4 -3/2 11 I'll just do one to show you how I figured out the y coordinates. 6 (x squared) + x - 1 = y 6 (1 squared) + 1 - 1 = y 6 (1) + 1 - 1 = y 6 + 1 - 1 = y 6 = y If you can see anything wrong or have any solutions to help me graph this I would really appreciate it. Brooke Date: 6/17/96 at 9:0:10 From: Doctor Anthony Subject: Re: Quadratic Problem >roots (-1/2, 0) (1/3, 0) The roots are (-1/2) and (1/3). Do NOT include the 0's >FINDING VERTEX You can find the vertex without first finding the roots. If the roots are irrational or non-existent, then you need an alternative approach anyhow. 6x^2 + x - 1 = y First make the coefficient of x^2 equal to 1 by dividing through by 6. x^2 + (1/6)x - 1/6 = y/6 Now complete the square on x by adding (and later subtracting) the square of half the coefficient of x. The coefficient of x is 1/6, so halve it = 1/12 and then square it = 1/144 x^2 + (1/6)x + 1/144 -1/6 - 1/144 = y/6 (x + 1/12)^2 - 25/144 = y/6 This reaches its minimum value when the bracket is zero (since anything squared will be positive). The bracket is zero if x = -1/12, and in this case y/6 = -25/144, so y = -25/24 Vertex is at (-1/12, -25/24) >FINDING COORDINATES If I were you, I would make up a table as shown below to plot the graph of the function: x| -2 | -1 | 0 | 1 | 2 | 3 | ---------------------------------------- 6x^2| 24 | 6 | 0 | 6 | 24 | 54 | x | -2 | -1 | 0 | 1 | 2 | 3 | -1 | -1 | -1 | -1 | -1 | -1 | -1 | ---------------------------------------- y= | 21 | 4 | -1 | 6 | 25 | 56 | You can of course plot at closer intervals if you want to, and for a different range of values of x, but a table layout as I have shown it above will make the calculations easier to follow. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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