1Cademy - Finding the Intercepts of $$y = -x^2 + 4x + 3$$

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Finding the Intercepts of a Parabola

Example

Finding the Intercepts of $y = -x^2 + 4x + 3$

To find the intercepts of the parabola $y = -x^2 + 4x + 3$ :

Finding the y-intercept: Let $x = 0$ and substitute it into the equation to solve for $y$ :

$y = -(0)^2 + 4(0) + 3$ $y = 3$

The y-intercept is the point $(0, 3)$ .

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

$0 = -x^2 + 4x + 3$

This quadratic does not factor, so use the Quadratic Formula with $a = -1$ , $b = 4$ , and $c = 3$ :

$x = \frac{-4 \pm \sqrt{4^2 - 4(-1)(3)}}{2(-1)}$

Simplify the expression:

$x = \frac{-4 \pm \sqrt{16 + 12}}{-2}$ $x = \frac{-4 \pm \sqrt{28}}{-2}$

Simplify the radical ( $\sqrt{28} = 2\sqrt{7}$ ):

$x = \frac{-4 \pm 2\sqrt{7}}{-2}$

Factor out $-2$ from the numerator to reduce:

$x = \frac{-2(2 \pm \sqrt{7})}{-2}$ $x = 2 \pm \sqrt{7}$

The x-intercepts are the exact points $(2 + \sqrt{7}, 0)$ and $(2 - \sqrt{7}, 0)$ . For graphing purposes, decimal approximations can be used: $(2 + \sqrt{7}, 0) \approx (4.6, 0)$ and $(2 - \sqrt{7}, 0) \approx (-0.6, 0)$ .

Updated 2026-04-21

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OpenStax Elementary Algebra

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