Q.Describe the shape of the graph of $y=x^2$, stating its vertex, axis of symmetry, and the intervals where it increases and decreases.

This is the basic quadratic polynomial, whose graph is a well-known curve shape that can be described from the algebraic form alone, without plotting individual points.

Step 1 — Overall shape. $y=x^2$ is a quadratic polynomial with leading coefficient $1>0$, so its graph is a parabola opening upward.

Step 2 — Vertex. Since $x^2 \ge 0$ for every real $x$, with equality only at $x=0$, the smallest value of $y$ is $y=0$, attained at $x=0$. The vertex is $(0,0)$.

Step 3 — Symmetry. Replacing $x$ by $-x$ gives $(-x)^2=x^2$, the same output — so the graph is unchanged under this reflection, meaning it is symmetric about the vertical line $x=0$ (the $y$-axis).

Step 4 — Increasing/decreasing behaviour. For two points with $x_1<x_2<0$: $x_1^2>x_2^2$ (both negative, but $x_1$ is more negative so its square is larger) — so $y$ decreases as $x$ increases toward $0$ from the left. For $0<x_1<x_2$: $x_1^2<x_2^2$ — so $y$ increases as $x$ increases beyond $0$. …

Confusing 'symmetric about the $y$-axis' with 'symmetric about the origin' — $y=x^2$ has the former, not the latter (an odd function like $y=x^3$ would have origin symmetry instead). Another slip is describing the curve as 'decre …