Intercepts and Symmetry Points of a Quadratic Function
Grade 9 students in California Reveal Math Algebra 1 learn to find the y-intercept, x-intercepts, and symmetry mirror points of a quadratic function in standard form f(x)=ax²+bx+c. The y-intercept is always (0,c). The x-intercepts are found by solving ax²+bx+c=0 — a parabola may have two, one, or no real zeros depending on the discriminant. Using the axis of symmetry x=-b/(2a), any known point gains a free mirror point at the same y-value on the opposite side. For example, f(x)=x²-4x+3 has y-intercept (0,3), x-intercepts (1,0) and (3,0), and axis x=2.
Key Concepts
For a quadratic function in standard form $f(x) = ax^2 + bx + c$:.
Y intercept: Set $x = 0$ to get the point $(0,\, c)$.
Common Questions
How do you find the y-intercept of a quadratic function?
Set x=0. For f(x)=ax²+bx+c, the y-intercept is always (0,c) — the constant term evaluated at x=0.
How do you find the x-intercepts of a quadratic function?
Set f(x)=0 and solve ax²+bx+c=0. The solutions are the x-values where the parabola crosses the x-axis.
How many x-intercepts can a parabola have?
A parabola can have 0, 1, or 2 real x-intercepts depending on the discriminant. If b²-4ac>0: two intercepts; if zero: one intercept (vertex touches x-axis); if negative: no real intercepts.
How do symmetry points help with graphing?
The axis of symmetry x=-b/(2a) lets you find a mirror point for any known point at no extra calculation. For example, the y-intercept (0,c) has a free mirror point at the same y-value on the other side of the axis.
Can you show a full example with f(x)=x²-4x+3?
y-intercept: (0,3). x-intercepts: solve x²-4x+3=0 → (x-1)(x-3)=0, giving (1,0) and (3,0). Axis of symmetry: x=4/2=2, confirming (1,0) and (3,0) are symmetric about x=2.
Which unit covers intercepts and symmetry of quadratics?
This skill is from Unit 10: Quadratic Functions in California Reveal Math Algebra 1, Grade 9.