Graphing Parabolas from Vertex Form

Write a quadratic in vertex form \(y = a(x-h)^2+k\) and its graph — a parabola — tells you almost everything at a glance: where it turns, which way it opens, and how wide it is.

By the end you'll be able to read the vertex, axis of symmetry, opening direction, and width straight off \(a(x-h)^2+k\) without plotting a single point.

Predict: if you drag a from 2 down to -2, does the vertex move? Drag the slider below to check.

This plots \(y = a(x-h)^2+k\). Drag a, h, and k and watch the vertex and the dashed axis of symmetry track it — the faint dotted curve is \(y=x^2\) for a width comparison. Hover the curve, or focus the vertex dot, for exact readouts.

Vertex (0.0, 0.0); axis \(x = 0.0\); opens upward (minimum); narrower than \(y=x^2\).

y = a(x−h)² + k — curve, vertex, axis of symmetry, and \(y=x^2\) for comparison
y = a(x−h)²+k vertex (h, k) axis x = h y = x² (reference)

Every quadratic's graph is a parabola, and vertex form \(y=a(x-h)^2+k\) puts its vertex, opening direction, and width right in the equation — no plotting required to see the shape.

Visual

Picture \(y=x^2\) — the plain U-shape sitting at the origin. Vertex form just moves and reshapes that same U: \(h\) slides it left/right, \(k\) slides it up/down, and \(a\) flips it (sign) and squeezes or stretches it (size) — the same shift-and-scale moves you've already seen for any function's graph. The vertex (h, k) is the turning point, and the axis of symmetry x = h is the mirror line the two arms fold across.

Formal

From \(y = a(x-h)^2+k\) you read off four features at once:

  • Vertex (h, k) — watch the sign: \((x-5)^2\) has \(h=5\), while \((x+5)^2 = (x-(-5))^2\) has \(h=-5\).
  • Axis of symmetry x = h — the vertical line the curve mirrors across.
  • Opening direction — the sign of a. \(a>0\) opens upward (vertex is a minimum); \(a<0\) opens downward (vertex is a maximum).
  • Width — the size of \(|a|\). Larger \(|a|\) makes a narrower, steeper curve; \(0 < |a| < 1\) makes it wider and flatter, compared to \(y=x^2\).

You get vertex form by completing the square on \(ax^2+bx+c\), and the axis \(x=h\) equals \(-b/(2a)\) in standard form.

Applied

A ball's height over time, a satellite dish's cross-section, a suspension cable under its own tension — all trace parabolas. Once you write the model in vertex form, the peak height (or the dish's focal point) is just the vertex, sitting right there in the equation, with no calculus or table of values needed to find it.

Worked example

Graph \(y = -2(x-3)^2+8\). Vertex \((h,k)=(3,8)\). Axis of symmetry \(x=3\). \(a=-2<0\), so it opens downward — the vertex \((3,8)\) is a maximum. \(|a|=2>1\), so it's narrower than \(y=x^2\). Symmetric point: at \(x=4\) (one unit right of the axis), \(y=-2(1)^2+8=6\), giving \((4,6)\); by symmetry \((2,6)\) is also on the curve. Setting \(y=0\) gives \(2(x-3)^2=8\), \((x-3)^2=4\), \(x-3=\pm2\), so x-intercepts at \(x=1\) and \(x=5\) — their average, 3, matches the axis, a handy self-check.

Your turn

For \(y = -3(x+2)^2+5\): vertex \((h,k) = (-2,5)\) — remember \((x+2)^2=(x-(-2))^2\). Axis of symmetry \(x=-2\). \(a=-3\), which is negative, so it opens ____ and the vertex is a ____.

Reveal the answer

\(a = -3 < 0\), so it opens downward, and the vertex is a maximum. Since \(|a| = 3 > 1\), it's also narrower than \(y = x^2\). Set a = -3, h = -2, k = 5 on the sliders above and check it against the vertex tooltip.

More info — vertex form as a shift and stretch of y = x²

Vertex form is exactly the shift/scale/reflect pattern from function transformations, applied to \(f(x)=x^2\): replacing \(x\) with \((x-h)\) shifts the graph right by \(h\); adding \(k\) shifts it up by \(k\); multiplying by \(a\) stretches it vertically by \(|a|\) and flips it when \(a<0\). If you'd rather derive vertex form from standard form \(ax^2+bx+c\), that's completing the square — the same algebraic move behind the quadratic formula. See the Paul's Online Notes link in Dive deeper for more worked sketches.

Check your understanding

Question 1 of 4

For \(y = 3(x + 1)^2 - 4\), give the vertex, axis of symmetry, and opening direction.

Question 2 of 4

Predict: starting from \(y = 2(x - 3)^2 + 1\), you drag a from 2 down to -2 (h and k unchanged). What happens to the vertex?

Question 3 of 4

Compared to \(y = x^2\), how does the graph of \(y = 0.25(x - 2)^2 - 1\) look, ignoring its shift?

Question 4 of 4

A parabola written in vertex form has vertex (2, 3) and opens upward. What does this say about its number of real x-intercepts, and how does that match the discriminant \(b^2-4ac\) from the quadratic formula?

Recap

  • A quadratic's graph is a parabola; vertex form is \(y=a(x-h)^2+k\).
  • Vertex \((h,k)\) is the turning point — watch the sign inside the parentheses: \((x+5)^2\) means \(h=-5\).
  • Axis of symmetry: \(x=h\), the mirror line the curve folds across.
  • Opening direction is the sign of a: \(a>0\) upward (minimum), \(a<0\) downward (maximum).
  • Width is set by \(|a|\): bigger \(|a|\) is narrower/steeper; \(0<|a|<1\) is wider/flatter than \(y=x^2\). Only \(h\) and \(k\) move the vertex — \(a\) never does.

Dive deeper

Sources

  • Graphing Parabolas from Vertex Form