Question

please use the answer in the attached picture to write down the co-ordinate of the turning point of x^2-5x+4

Answer 1

Solution

By general rule w ehave this property:

$(x-a)^2=x^2-2ax+a^2$

Using this expression we can do the following:

$(x-\frac{5}{2})^2-\frac{9}{4}=(x\cdot x^{}-2\cdot x\cdot\frac{5}{2}+\frac{5}{2}\cdot\frac{5}{2})-\frac{9}{4}=x^2-5x+\frac{25}{4}-\frac{9}{4}=x^2-5x+\frac{16}{4}=x^2-5x+4=(x-1)(x-4)$

Then the final simplfied form is:

(x-1)(x-4)

Then we can find the turning point like this:

$x=-\frac{b}{2a}=-\frac{-5}{2\cdot1}=\frac{5}{2}$

And the corresponding y coordinate is:

$(\frac{5}{2}-\frac{5}{2})^2-\frac{9}{4}=-\frac{9}{4}$

Final answer:

$(\frac{5}{2},-\frac{9}{4})$