This problem is about identifying x-intercepts, y intercepts, the vertex, the axis of quadratics and how to graph it. Here's an example where I used my own example.
To begin with, my problem was f(x)=2x^2+4x-16. I subtracted 16 on both sides which gave me 2x^2+4x=16. I then took out the coeficiant which is 2 and added it to the other side as well, and it gave me 2(x^2+2x)=16+2. After, I divided "b" by 2 which is 1 and squared it which is 1 and added 1 to both sides. I ended up with 2(x^2+2x+1)= 16+2(1). KEEP IN MIND THAT "2(x^2+2x+1)" FACTORS OUT. 2(x+1)^2=18. Subtract 18 over and we have the parent function equation which is 2(x+1)^2-18. The vertex is a minimum and is (-1,-18). To find the y intercept, we would need to set "x"=to 0 in the original equation which makes it (0,-16). The axis is "x=-1" because of the vertex. Now to find the x intercepts, we just need to solve, "2(x+1)^2=18" :)
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