Question

Evaluate h(x) = 2.3x^3 + 0.01x^2 for x=1 and x=2

Answer 1

Step-by-step explanation

For the expression

`h(x)=2.3x^3+0.01x^2`

To find the value of h(x) when x=1

we will put x =1 into the expression

`\begin{gathered} h(x)=2.3(1)+0.01(1) \\ h(1)=2.3+0.01 \\ h(1)=2.31 \end{gathered}`

When x=2

we put x=2 into the expression

`\begin{gathered} h(2)=2.3*(2)^3+0.01*(2)^2 \\ h(2)=18.4+0.04 \\ h(2)=18.44 \end{gathered}`