1Cademy - Graphing $$y = 2x^2

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Graphing a Quadratic Equation in Two Variables

Example

Graphing $y = 2x^2 - 6x + 5$

To graph the parabola $y = 2x^2 - 6x + 5$ , follow the standard graphing procedure:

Step 1: The coefficient $a = 2$ is positive, so the parabola opens upward.

Step 2: Find the axis of symmetry using $x = -\frac{b}{2a}$ . Since $a = 2$ and $b = -6$ , $x = -\frac{-6}{2(2)} = \frac{6}{4} = \frac{3}{2}$ . The axis of symmetry is the line $x = \frac{3}{2}$ .

Step 3: Find the vertex by substituting $x = \frac{3}{2}$ into the equation: $y = 2(\frac{3}{2})^2 - 6(\frac{3}{2}) + 5 = 2(\frac{9}{4}) - 9 + 5 = \frac{9}{2} - 4 = \frac{1}{2}$ . The vertex is the point $(\frac{3}{2}, \frac{1}{2})$ .

Step 4: Find the $y$ -intercept by setting $x = 0$ : $y = 2(0)^2 - 6(0) + 5 = 5$ . The $y$ -intercept is $(0, 5)$ . The point symmetric to the $y$ -intercept across the axis of symmetry $x = \frac{3}{2}$ is $(3, 5)$ .

Step 5: Find the $x$ -intercepts by setting $y = 0$ , yielding $0 = 2x^2 - 6x + 5$ . Testing the discriminant gives $b^2 - 4ac = (-6)^2 - 4(2)(5) = 36 - 40 = -4$ . Because the discriminant is negative, there are no real solutions, meaning the parabola has no $x$ -intercepts.

Step 6: Graph the parabola by plotting the vertex, the $y$ -intercept, and the symmetric point, then connecting them with a smooth, upward-opening curve.

Updated 2026-04-21

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OpenStax Elementary Algebra

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