Question

Vertex of x^2-10x+30

Answer 1

We have: $f(x)=x^2-10x+30$

Method 1.

Use: $(*)\ \ \ \ \ (a-b)^2=a^2-2ab+b^2$

and the vertex form $f(x)=a(x-h)^2+k$

(h, k) - the coordinates of a vertex

$x^2-10x+30=x^2-2\cdot x\cdot5+30=\underbrace{x^2-2\cdot x\cdot5+5^2}_{(*)}-5^2+30\\\\=(x-5)^2+5\to h=5;\ k=5$

Answer: (5, 5)

Method 2.

For $f(x)=ax^2+bx+c$ the coordinates of a vertex (h, k) are equal

$h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}$

We have: $f(x)=x^2-10x+30\to a=1;\ b=-10;\ c=30$

Substitute:

$h=\dfrac{-(-10)}{2\cdot1}=\dfrac{10}{2}=5\\\\k=f(5)=5^2-10\cdot5+30=25-50+30=5$

Answer: (5, 5)

Answer 2

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).