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Ksivusya [100]
3 years ago
15

Choose the correct vertex of the function f(x) = x2 - X + 2.

Mathematics
1 answer:
lys-0071 [83]3 years ago
3 0

Answer:

\large\boxed{\left(\dfrac{1}{2},\ \dfrac{7}{4}\right)}

Step-by-step explanation:

The vertex form of an equation of a parabola <em>f(x) = ax² + bx + c:</em>

f(x)=a(x-h)^2+k

<em>(h, k)</em><em> - vertex</em>

<h3>METHOD 1:</h3>

h=\dfrac{-b}{2a},\ k=f(h)\\\\f(x)=x^2-x+2\to a=1,\ b=-1,\ c=2\\\\h=\dfrac{-(-1)}{(2)(1)}=\dfrac{1}{2}\\\\k=f\bigg(\dfrac{1}{2}\bigg)=\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}+2=\dfrac{1}{4}-\dfrac{1}{2}+2=\dfrac{1}{4}-\dfrac{1\cdot2}{2\cdot2}+2\\\\=\dfrac{1}{4}-\dfrac{2}{4}+2=-\dfrac{1}{4}+2=1\dfrac{3}{4}=\dfrac{7}{4}\\\\\boxed{\left(\dfrac{1}{2},\ \dfrac{7}{4}\right)}

<h3>METHOD 2:</h3>

\text{use}\ (a-b)^2=a^2-2ab+b^2\qquad(*)\\\\f(x)=x^2-x+2=\underbrace{x^2-2(x)\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2}_{(*)}-\left(\dfrac{1}{2}\right)^2+2\\\\=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+2=\left(x-\dfrac{1}{2}\right)^2+1\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\\\\h=\dfrac{1}{2},\ k=\dfrac{7}{4}\\\\\boxed{\left(\dfrac{1}{2},\ \dfrac{7}{4}\right)}

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