Parabola Vertex Calculator

Enter the parabola equation to calculate its vertex, determine if it is a maximum or minimum point, and see the detailed step-by-step resolution along with its graph.

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Solved Exercises

Calculate the vertex of the parabola $ y=2x^2-8x+5 $ and determine if it is a maximum or a minimum.

Result

The vertex of the given parabola is:

$$ V\left(2, -3\right) $$

It is the minimum point of the curve, since the parabola opens upward.

Step-by-step resolution

The given equation is:

$$ y = 2 x^{2}-8 x+5 $$

1. Identification of the form and coefficients

The equation is in its standard form $ y = ax^2 + bx + c $ and corresponds to a vertical parabola (because the quadratic variable is x). We extract the corresponding coefficients:

$$ a = 2 \quad b = -8 \quad c = 5 $$

2. Calculation of the vertex x-coordinate (h)

To find the vertex x-coordinate, we apply the following formula:

$$ h = \dfrac{-b}{2a} \\[1em] h = \dfrac{-\left(-8\right)}{2\left(2\right)} \\[1em] h = 2 $$

3. Calculation of the vertex y-coordinate (k)

To find the y-coordinate, we evaluate the h coordinate found in the original equation:

$$ k = 2\left(2\right)^2 - 8\left(2\right) + 5 \\[1em] k = -3 $$

4. Vertex and its properties

With the obtained coordinates, we know that the vertex of the parabola is located at:

$$ V\left(2, -3\right) $$

Since $ a = 2 > 0 $, the parabola opens upward, therefore the vertex is the minimum point. Additionally, the axis of symmetry of the parabola passes through the vertex x-coordinate; its equation is:

$$ x = 2 $$

Graph in the Cartesian plane of a vertical parabola opening upward and its vertex outside the origin. — Graph of the parabola and its vertex

Find the vertex coordinates of the following parabola: $ y=-3(x+4)^2+7. $

Result

The vertex of the given parabola is:

$$ V\left(-4, 7\right) $$

It is the maximum point of the curve, since the parabola opens downward.

Step-by-step resolution

The given equation is:

$$ y = -3\left(x + 4\right)^2 + 7 $$

1. Identification of the form and coefficients

The equation is in its vertex form $ y = a(x - h)^2 + k $, where the vertex is (h, k). This form allows us to extract the coordinates of the vertex directly without the need to apply formulas.

2. Extraction of values from the equation

By comparing the equation with the generic vertex form, we extract directly the h and k vertex coordinates and the coefficient a:

$$ h = -4 $$

$$ k = 7 $$

$$ a = -3 $$

4. Vertex and its properties

With the obtained coordinates, we know that the vertex of the parabola is located at:

$$ V\left(-4, 7\right) $$

Since $ a = -3 < 0 $, the parabola opens downward, therefore the vertex is the maximum point. Additionally, the axis of symmetry of the parabola passes through the vertex x-coordinate; its equation is:

$$ x = -4 $$

Graph in the Cartesian plane of a vertical parabola opening downward and its vertex outside the origin. — Graph of the parabola with its vertex

Obtain the vertex coordinates for the horizontal parabola defined by $ x=0.5y^2+2y-1. $

Solution

The vertex of the given parabola is:

$$ V\left(-3, -2\right) $$

It is the point with the minimum x-coordinate of the curve (leftmost point), since it opens to the right.

Step-by-step resolution

The given equation is equivalent to:

$$ x = \dfrac{y^{2}}{2}+2 y-1 $$

1. Identification of the form and coefficients

The equation is in its standard form $ x = ay^2 + by + c $ and corresponds to a horizontal parabola (because the quadratic variable is y). We extract the corresponding coefficients:

$$ a = \dfrac{1}{2} \quad b = 2 \quad c = -1 $$

2. Calculation of the vertex y-coordinate (k)

To find the vertex y-coordinate, we apply the following formula:

$$ k = \dfrac{-b}{2a} \\[1em] k = \dfrac{-2}{2\left(\dfrac{1}{2}\right)} \\[1em] k = -2 $$

3. Calculation of the vertex x-coordinate (h)

To find the x-coordinate, we evaluate the k coordinate found in the original equation:

$$ h = \dfrac{1}{2}\left(-2\right)^2 + 2\left(-2\right) - 1 \\[1em] h = -3 $$

4. Vertex and its properties

With the obtained coordinates, we know that the vertex of the parabola is located at:

$$ V\left(-3, -2\right) $$

Since $ a = \dfrac{1}{2} > 0 $, the parabola opens to the right, therefore the vertex is the point with the minimum x-coordinate. Additionally, the axis of symmetry of the parabola passes through the vertex y-coordinate; its equation is:

$$ y = -2 $$

Graph in the Cartesian plane of a horizontal parabola opening to the right and its vertex outside the origin. — Graph of the parabola and its vertex

Calculate the vertex of the parabola given by the quadratic function $ f(x)=x^2+6x. $

Answer

The vertex of the given parabola is:

$$ V\left(-3, -9\right) $$

It is the minimum point of the curve, since the parabola opens upward.

Step-by-step resolution

The given equation is equivalent to:

$$ y = x^{2}+6 x $$

1. Identification of the form and coefficients

The equation is in its standard form $ y = ax^2 + bx + c $ and corresponds to a vertical parabola (because the quadratic variable is x). We extract the corresponding coefficients:

$$ a = 1 \quad b = 6 \quad c = 0 $$

2. Calculation of the vertex x-coordinate (h)

To find the vertex x-coordinate, we apply the following formula:

$$ h = \dfrac{-b}{2a} \\[1em] h = \dfrac{-6}{2\left(1\right)} \\[1em] h = -3 $$

3. Calculation of the vertex y-coordinate (k)

To find the y-coordinate, we evaluate the h coordinate found in the original equation:

$$ k = \left(-3\right)^2 + 6\left(-3\right) \\[1em] k = -9 $$

4. Vertex and its properties

With the obtained coordinates, we know that the vertex of the parabola is located at:

$$ V\left(-3, -9\right) $$

Since $ a = 1 > 0 $, the parabola opens upward, therefore the vertex is the minimum point. Additionally, the axis of symmetry of the parabola passes through the vertex x-coordinate; its equation is:

$$ x = -3 $$

Graph in the Cartesian plane of a quadratic function and its vertex outside the origin as a minimum point. — Graph of the parabola and its vertex

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Professor of Mathematics, graduated from the Faculty of Exact, Chemical and Natural Sciences of the National University of Misiones (UNAM). Developer and creator of RigelUp, dedicated to building tools for mathematical learning.