Question

sketch the graph of a function with the following features:

1. the y - intercept is 1
2. the function is positive for all values of x
3. the function is increasing for x > 1
4. as x → ∞, f(x) → ∞ and as x → - ∞, f(x) → ∞

Explanation:

Step1: Consider the y - intercept

Since the y - intercept is \(1\), the point \((0,1)\) is on the graph.

Step2: Analyze the sign of the function

The function \(y = f(x)>0\) for all \(x\in R\).

Step3: Analyze the increasing - decreasing behavior

The function is increasing for \(x > 1\). Also, since \(\lim_{x
ightarrow\infty}f(x)=\infty\) and \(\lim_{x
ightarrow-\infty}f(x)=\infty\), we can consider a quadratic - like function (but adjusted for the given conditions). A simple example is \(y=(x - 1)^2+1=x^{2}-2x + 2\).

• For the y - intercept: when \(x = 0\), \(y=(0 - 1)^2+1=1\).
• For the sign: \(y=(x - 1)^2+1\), since \((x - 1)^2\geq0\) for all \(x\in R\), then \(y=(x - 1)^2+1\geq1>0\) for all \(x\in R\).
• For the increasing - decreasing: take the derivative \(y^\prime=2(x - 1)\). When \(x>1\), \(y^\prime=2(x - 1)>0\) (function is increasing). And \(\lim_{x

ightarrow\pm\infty}(x^{2}-2x + 2)=\lim_{x
ightarrow\pm\infty}x^{2}(1-\frac{2}{x}+\frac{2}{x^{2}})=\infty\)

Answer:

A possible graph is of the function \(y=(x - 1)^2+1\). Plot the point \((0,1)\). The parabola opens upwards (since the leading coefficient of \(x^{2}-2x + 2\) is positive). For \(x>1\), the function is increasing. The function is always above the \(x\) - axis.