Example

Finding the Intercepts of y=5x2+x+4y = 5x^2 + x + 4

To find the intercepts of the parabola y=5x2+x+4y = 5x^2 + x + 4:

Finding the y-intercept: Let x=0x = 0 and solve for yy:

y=5(0)2+0+4=4y = 5(0)^2 + 0 + 4 = 4

When x=0x = 0, y=4y = 4. The y-intercept is the point (0,4)(0, 4).

Finding the x-intercepts: Let y=0y = 0 and solve for xx:

0=5x2+x+40 = 5x^2 + x + 4

Before attempting to solve, compute the discriminant to predict how many solutions exist. Identify a=5a = 5, b=1b = 1, c=4c = 4:

b24ac=12454=180=79b^2 - 4ac = 1^2 - 4 \cdot 5 \cdot 4 = 1 - 80 = -79

Because the discriminant is negative, the quadratic equation has no real solutions. Therefore, the parabola has no x-intercepts — it does not cross the x-axis.

This example demonstrates a case where using the discriminant saves time: rather than applying the full Quadratic Formula, the negative discriminant immediately reveals that no real x-values make y=0y = 0, so the only intercept is the y-intercept (0,4)(0, 4).

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Updated 2026-04-21

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