Graph f(x) = x2 + 1, giving its vertex, axis of symmetry and range.

Step-by-step solution

1. 1

   Recognise the parent function and the transformation. Written in vertex form $a(x-h)^2+k$, the function is $1\cdot(x-0)^2+1$, so $a=1$, $h=0$, $k=1$. That means: take the parent parabola $y=x^2$ and shift it up 1 unit. Nothing is stretched or reflected, since $a=1$.

2. 2

   Read off the vertex, axis and direction. The vertex is $(h,k)=(0,1)$ and the axis of symmetry is the vertical line $x=h=0$ (the $y$-axis). Because $a=1>0$ the parabola opens upward, so the vertex is a minimum.

3. 3

   Build a small symmetric table of values.

   $f(-2)=5,\quad f(-1)=2,\quad f(0)=1,\quad f(1)=2,\quad f(2)=5$

   The repeated $y$-values in mirrored pairs ($\pm1\mapsto 2$, $\pm2\mapsto 5$) are a direct check that the axis really is $x=0$.

4. 4

   Check the intercepts. Setting $x=0$ gives the $y$-intercept $(0,1)$. For $x$-intercepts, $x^2+1=0$ needs $x^2=-1$, which has no real solution — so the curve never crosses the $x$-axis. That is exactly what the minimum value $1>0$ predicts.

5. 5

   Describe the finished graph. Plot $(0,1)$, then $(\pm1,2)$ and $(\pm2,5)$, and join them in a smooth U opening upward, the same width as $y=x^2$. Domain: all real numbers. Range: $y\ge 1$, i.e. $[1,\infty)$.