Identifying the Y-Intercept from Standard Form
Identifying the y-intercept from standard form is a Grade 11 Algebra 1 skill from enVision Chapter 8 showing that the constant term c in f(x) = ax^2 + bx + c directly gives the y-intercept at (0, c). Substituting x = 0 eliminates both the ax^2 and bx terms, leaving f(0) = c. For f(x) = 2x^2 - 5x + 3, the y-intercept is (0, 3). For g(x) = -x^2 + 4x - 7, it is (0, -7). When no constant term exists, as in h(x) = 3x^2 + 2x, the y-intercept is (0, 0), meaning the parabola passes through the origin.
Key Concepts
In the standard form of a quadratic function $f(x) = ax^2 + bx + c$, the constant term $c$ represents the y intercept of the parabola. The y intercept occurs at the point $(0, c)$.
Common Questions
How do you find the y-intercept of a quadratic in standard form?
The constant term c in f(x) = ax^2 + bx + c is the y-intercept. The y-intercept is the point (0, c).
Why does substituting x = 0 give the y-intercept?
At x = 0, f(0) = a(0)^2 + b(0) + c = c. The quadratic and linear terms vanish, leaving only the constant.
What is the y-intercept of f(x) = 2x^2 - 5x + 3?
c = 3, so the y-intercept is (0, 3).
What is the y-intercept of g(x) = -x^2 + 4x - 7?
c = -7, so the y-intercept is (0, -7). The parabola crosses the y-axis below the origin.
What if there is no constant term, like h(x) = 3x^2 + 2x?
c = 0, so the y-intercept is (0, 0). The parabola passes through the origin.
Is the y-intercept always a single point for a quadratic function?
Yes. A function has exactly one y-intercept because a function assigns exactly one output to each input, including x = 0.