1Cademy - Graphing $$y = 25x^2 + 10x + 1$$

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Graphing a Quadratic Equation in Two Variables

Example

Graphing $y = 25x^2 + 10x + 1$

To graph the parabola $y = 25x^2 + 10x + 1$ , apply the standard graphing procedure:

Step 1: The coefficient $a = 25$ is positive, so the parabola opens upward.

Step 2: Find the axis of symmetry using $x = -\frac{b}{2a}$ . Since $b = 10$ and $a = 25$ , $x = -\frac{10}{2(25)} = -\frac{10}{50} = -\frac{1}{5}$ . The axis of symmetry is the line $x = -\frac{1}{5}$ .

Step 3: Find the vertex by substituting $x = -\frac{1}{5}$ into the equation: $y = 25(-\frac{1}{5})^2 + 10(-\frac{1}{5}) + 1 = 25(\frac{1}{25}) - 2 + 1 = 1 - 2 + 1 = 0$ . The vertex is the point $(-\frac{1}{5}, 0)$ .

Step 4: Find the $y$ -intercept by setting $x = 0$ : $y = 25(0)^2 + 10(0) + 1 = 1$ . The $y$ -intercept is $(0, 1)$ . The point symmetric to the $y$ -intercept across the axis of symmetry $x = -\frac{1}{5}$ is $(-\frac{2}{5}, 1)$ .

Step 5: Find the $x$ -intercepts by setting $y = 0$ : $0 = 25x^2 + 10x + 1$ . This factors as a perfect square trinomial: $0 = (5x + 1)^2$ . Solving yields $5x + 1 = 0$ , so $x = -\frac{1}{5}$ . The only $x$ -intercept is $(-\frac{1}{5}, 0)$ , which coincides with the vertex.

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

Updated 2026-04-21

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OpenStax Elementary Algebra

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