Question

If $p(𝑥)=x^2-4𝑥+3$, evaluate: $𝑝\left(2\right)-𝑝(-1)+𝑝\left(\frac{1}{2}\right)$.

Answer 1

Find the value of $p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right)$

$p\left(x\right)=x^2–4x+3$

According to the given details

When $x=2$, $p\left(x\right)=p\left(2\right)$

Substituting $x=2$,

$$\begin{gathered} p\left(2\right)={\left(2\right)}^2–4\left(2\right)+3 \\ =4-8+3 \\ =-4+3 \\ =–1 \end{gathered}$$

When $x=–1$, $p\left(x\right)=p(–1)$

Substituting $x=-1$,

$$\begin{gathered} p(–1)={(–1)}^2-4(–1)+3 \\ =1+4+3 \\ =8 \end{gathered}$$

When $x=\frac{1}{2}$ , $p\left(x\right)=p\left(\frac{1}{2}\right)$

Substituting $x=\frac{1}{2}$,

$$\begin{gathered} p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-4\left(\frac{1}{2}\right)+3 \\ =\left(\frac{1}{4}\right)-2+3 \\ =\left(\frac{1}{4}\right)+1 \\ =\frac{5}{4} \end{gathered}$$

Now substitute values in given equation

$$\begin{gathered} p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right)=–1–8+\left(\frac{5}{4}\right) \\ =-9+\frac{5}{4} \\ =\frac{-36+5}{4} \\ =\frac{-31}{4} \end{gathered}$$

Hence the value of $p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right)$ is $-\frac{31}{4}$