The easiest way to find the vertex is to convert this standard form equation into vertex form, which is y = a(x - h)^2 + k.
Firstly, put x^2 - 10x into parentheses: y = (x^2 - 10x) + 30
Next, we want to make what's inside the parentheses a perfect square. To do that, we need to divide the x coefficient by 2 and square it. In this case, the result is 25. Add 25 inside the parentheses and subtract 25 outside of the parentheses: y = (x^2 - 10x + 25) + 30 - 25
Next, factor what's inside the parentheses and combine like terms outside of the parentheses, and your vertex form is: y = (x - 5)^2 + 5.
Now going back to the formula of the vertex form, y = a(x - h)^2 + k, the vertex is (h,k). Using our vertex equation, we can see that the vertex is (5,5).
We have: [tex]f(x)=x^2-10x+30[/tex]
Method 1.
Use: [tex](*)\ \ \ \ \ (a-b)^2=a^2-2ab+b^2[/tex]
and the vertex form [tex]f(x)=a(x-h)^2+k[/tex]
(h, k) - the coordinates of a vertex
[tex] x^2-10x+30=x^2-2\cdot x\cdot5+30=\underbrace{x^2-2\cdot x\cdot5+5^2}_{(*)}-5^2+30\\\\=(x-5)^2+5\to h=5;\ k=5 [/tex]
Answer: (5, 5)
Method 2.
For [tex]f(x)=ax^2+bx+c[/tex] the coordinates of a vertex (h, k) are equal
[tex]h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}[/tex]
We have: [tex]f(x)=x^2-10x+30\to a=1;\ b=-10;\ c=30[/tex]
Substitute:
[tex]h=\dfrac{-(-10)}{2\cdot1}=\dfrac{10}{2}=5\\\\k=f(5)=5^2-10\cdot5+30=25-50+30=5[/tex]
Answer: (5, 5)
