1Cademy - Graphing $$y = 5x^2 + 10x + 3$$

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Graphing a Quadratic Equation in Two Variables

Example

Graphing $y = 5x^2 + 10x + 3$

To graph the parabola $y = 5x^2 + 10x + 3$ , follow the standard graphing procedure:

Step 1: The coefficient $a = 5$ is positive, so the parabola opens upward.

Step 2: Find the axis of symmetry using $x = -\frac{b}{2a}$ . With $a = 5$ and $b = 10$ , $x = -\frac{10}{2(5)} = -\frac{10}{10} = -1$ . The axis of symmetry is the line $x = -1$ .

Step 3: Find the vertex by substituting $x = -1$ into the equation: $y = 5(-1)^2 + 10(-1) + 3 = 5 - 10 + 3 = -2$ . The vertex is the point $(-1, -2)$ .

Step 4: Find the $y$ -intercept by setting $x = 0$ : $y = 5(0)^2 + 10(0) + 3 = 3$ . The $y$ -intercept is (0, 3). The point symmetric to the $y$ -intercept across the axis of symmetry $x = -1$ is $(-2, 3)$ .

Step 5: Find the $x$ -intercepts by setting $y = 0$ : $0 = 5x^2 + 10x + 3$ . Use the Quadratic Formula with $a = 5$ , $b = 10$ , and $c = 3$ :

$x = \frac{-10 \pm \sqrt{10^2 - 4(5)(3)}}{2(5)}$ $x = \frac{-10 \pm \sqrt{100 - 60}}{10}$ $x = \frac{-10 \pm \sqrt{40}}{10}$ $x = \frac{-10 \pm 2\sqrt{10}}{10}$

Factor out $2$ from the numerator to simplify:

$x = \frac{2(-5 \pm \sqrt{10})}{10} = \frac{-5 \pm \sqrt{10}}{5}$

The approximate values of the $x$ -intercepts are $(-0.4, 0)$ and $(-1.6, 0)$ .

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

Updated 2026-04-21

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OpenStax Elementary Algebra

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