= From this point, I began to square root both sides and remembered to put the sign on the right: x.
Once I did that, I solved for x and simplified as necessary: x=
By evaluating the answer, it finally resulted in the Quadratic Formula:
Problem 2: 2x?-x+2=0 Rewriting in vertex form: Before I rewrote the quadratic equation of 2x?-x+2=0 in vertex form, I had to understand that the vertex form of a quadratic function is f(x)=a(x-h)?+k where …show more content…
A parabola will always have a domain of all x-values so I immediately knew that the domain of this quadratic equation is all real numbers, or (?∞,∞). To find the range, I evaluated the (h,k) which in this equation is (1/4, 15/8). I can see that this parabola has a minimal value of 15/8 and go up to positive infinity. The range is y is greater than or equal to 15/8, or y. The maximum/minimum value can be found by the output of the quadratic function at the vertex. By evaluating the vertex in y=2(x-1/4)?+15/8, I noticed that -1/4 was a negative number which meant that it had a minimum value occurring at