Quadratic Vertex Form: y = a(x − h)² + k

The vertex form of a quadratic is the friendliest one:

$y = a(x - h)^{2} + k$

The constants have direct geometric meaning:

• $(h, k)$ is the vertex of the parabola — the highest or lowest point.
• $a$ controls the width and direction: positive $a$ opens upward, negative opens downward, large $|a|$ makes a narrow parabola, small $|a|$ makes it wide.

Compare this to the standard form $y = ax^{2} + bx + c$ — same parabola, but the vertex is hidden inside an ugly formula $\bigl(-b/(2a),\, c - b^{2}/(4a)\bigr)$. The vertex form makes it explicit.

Step-by-Step Solution

Given: $y = 2(x - 3)^{2} + 1$.

Find: The vertex, the direction the parabola opens, and the values at $x = 0, 3, 5$.

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Step 1 — Read off the vertex.

Compare $y = 2(x - 3)^{2} + 1$ to $y = a(x - h)^{2} + k$:

$a = 2,\;\; h = 3,\;\; k = 1$

The vertex is $\boxed{(h, k) = (3, 1)}$.

Step 2 — Determine the direction.

Since $a = 2 > 0$, the parabola opens upward. The vertex is therefore the minimum point.

Step 3 — Compute $y$ at $x = 0$.

$y(0) = 2(0 - 3)^{2} + 1 = 2(9) + 1 = 19$

Step 4 — Compute $y$ at $x = 3$ (the vertex itself).

$y(3) = 2(3 - 3)^{2} + 1 = 2(0) + 1 = 1$

The minimum value of $y$ on this parabola is 1, attained at the vertex.

Step 5 — Compute $y$ at $x = 5$.

$y(5) = 2(5 - 3)^{2} + 1 = 2(4) + 1 = 9$

Step 6 — How $h$ shifts the parabola.

Subtracting $h$ inside the squared term shifts the parabola right by $h$ (counterintuitive — like with phase shifts in trig functions). For the same parabola but with $h = 0$, the vertex would be at $(0, 1)$. For $h = 5$, vertex at $(5, 1)$.

Step 7 — How $k$ shifts the parabola.

Adding $k$ outside shifts the parabola up by $k$. With $k = -2$, the vertex moves to $(3, -2)$.

Step 8 — How $a$ stretches it.

For $a = 1$, the parabola has the same shape as $y = x^{2}$. For $a = 2$, it's twice as steep — more narrow. For $a = 0.5$, it's half as steep — wider. For $a < 0$, it opens downward instead of upward.

Step 9 — Convert to standard form.

Expand the vertex form:

$y = 2(x - 3)^{2} + 1 = 2(x^{2} - 6x + 9) + 1 = 2x^{2} - 12x + 18 + 1 = 2x^{2} - 12x + 19$

So in standard form, $y = 2x^{2} - 12x + 19$, with $a = 2$, $b = -12$, $c = 19$. The vertex from this form: $-b/(2a) = 12/4 = 3$, and $y(3) = 1$ — same answer, just hidden behind more algebra.

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Answer:

For $y = 2(x - 3)^{2} + 1$:

• Vertex: $\boxed{(3, 1)}$
• Direction: opens upward ($a > 0$)
• Minimum value: $y = 1$ at $x = 3$
• $y(0) = 19$, $y(5) = 9$, $y(3) = 1$ (the minimum)
• Standard form: $y = 2x^{2} - 12x + 19$

The vertex form makes the geometry obvious; the standard form requires extra algebra to extract the vertex.

Try It

• Adjust the vertex $(h, k)$ with the sliders.
• Adjust $a$ to stretch or flip the parabola.
• The vertex marker (yellow) tracks your changes in real time.
• Notice that $h$ shifts horizontally and $k$ shifts vertically — clean and intuitive.