Example

Finding the Intercepts of y=x2+4x+3y = x^2 + 4x + 3

To find the intercepts of the parabola y=x2+4x+3y = x^2 + 4x + 3:

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

y=02+4(0)+3y = 0^2 + 4(0) + 3 y=3y = 3

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

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

0=x2+4x+30 = x^2 + 4x + 3

Factor the quadratic equation:

0=(x+1)(x+3)0 = (x + 1)(x + 3)

Use the zero product property to set each factor equal to zero:

x+1=0orx+3=0x + 1 = 0 \quad \text{or} \quad x + 3 = 0

Solve the equations:

x=1orx=3x = -1 \quad \text{or} \quad x = -3

The x-intercepts are the points (1,0)(-1, 0) and (3,0)(-3, 0).

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Updated 2026-04-21

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