If $b = 0$, the x-coordinate of the vertex is $x = 0$. The y-coordinate is simply the constant $c$, making the vertex $(0, c)$.

Can the vertex formula be used for horizontal parabolas given by $x = ay^2 + by + c$?

Yes, but the variables are swapped. The y-coordinate of the vertex is $y = -\frac{b}{2a}$, and you substitute this into the equation to find the corresponding x-coordinate.

What if the coefficient $a$ is a fraction?

The formula $x = -\frac{b}{2a}$ applies exactly the same way. Multiply the fraction by 2 in the denominator, then divide $b$ by the resulting fraction.

How to find the vertex of a parabola from standard form

The vertex of a parabola given in standard form $y = ax^2 + bx + c$ is found using the vertex formula. The x-coordinate is $x = -\frac{b}{2a}$ , and the y-coordinate is found by substituting this x-value back into the original equation.

This method applies to any quadratic function where $a eq 0$ . It is the most direct approach when the equation is expanded, avoiding the need to complete the square.

The setup

Identify the coefficients $a$ , $b$ , and $c$ from the standard form equation $y = ax^2 + bx + c$ . Ensure the equation is ordered by descending powers of $x$ before extracting these values.

The steps

Calculate the x-coordinate of the vertex using $h = -\frac{b}{2a}$ .
Substitute $h$ into the original equation to find the y-coordinate: $k = a(h)^2 + b(h) + c$ .
Write the vertex as the coordinate pair $(h, k)$ .

Checking the result

Verify the result by completing the square to convert the equation into vertex form $y = a(x-h)^2 + k$ . Alternatively, check points equidistant from the vertex x-coordinate; $x = h - 1$ and $x = h + 1$ must yield identical y-values.

Common errors

Dropping the negative sign in the formula $x = -\frac{b}{2a}$ . Incorrectly squaring a negative $h$ value when evaluating $k$ (e.g., treating $-3^2$ as $-9$ instead of $9$ ). Misidentifying $b$ and $c$ if the terms are not in descending order.

Worked example

Find the vertex of the parabola given by $y = -2x^2 + 8x - 5$ .

Identify the coefficients: $a = -2$ $b = 8$ $c = -5$

Calculate the x-coordinate: $h = -\frac{b}{2a}$ $h = -\frac{8}{2(-2)}$ $h = -\frac{8}{-4}$ $h = 2$

Calculate the y-coordinate by substituting $x = 2$ : $k = -2(2)^2 + 8(2) - 5$ $k = -2(4) + 16 - 5$ $k = -8 + 16 - 5$ $k = 3$

The vertex is $(2, 3)$ .

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References: OpenStax College Algebra, Chapter 5: Quadratic Functions · Khan Academy: Quadratic equations & functions unit