Question

How do you find the x intercept and coordinates of the vertexfor the parabola y=x^2-14x+49????

Answer 1

$y=x^2-14x+49=\underbrace{x^2-2x\cdot7+7^2}_{use\ (a-b)^2=a^2-2ab+b^2}=(x-7)^2\\\\Vertex\ form\ of\ a\ quadratic\ function:\\\\y=a(x-h)^2+k\ where\ coordinates\ of\ the\ vertex\ (h;\ k)\\\\\boxed{vertex:(7;\ 0)}\\\\x-intercept\ (y=0):\\\\(x-7)^2=0\iff x-7=0\Rightarrow \boxed{x=7}\\\\\\Answer:\\\\x-intercept:(7;\ 0)\\\\the\ coordinates\ of\ the\ vertex:(7;\ 0)$

Answer 2

Square of (x) - 14x + 49 equals square of (x-7). Because there is a 49 and middle element of function is -14. If you use 49 with -7*-7, it means -7 + (-7) = -14 and -7*-7=49 so there is no problem.I said it's square of (x-7) it means (x-7)(x-7) there are 2 same roots and it's 7. If you want to find vertex of this parabola you must use -b/2a. It means 14/2*1 = 7 . x value of vertex is 7 , and if you put this number to x you will find it's y value and it's 0.