Q. Quadratic function y-intercept formula.

Answer

Set \(x = 0\).

Standard form:
\[
y = ax^2 + bx + c
\]
\(y(0) = c\), so the y-intercept is \((0,c)\).

Vertex form:
\[
y = a(x-h)^2 + k
\]
\(y(0) = a h^2 + k\), so the y-intercept is \((0, a h^2 + k)\).

Factored form:
\[
y = a(x – r_1)(x – r_2)
\]
\(y(0) = a r_1 r_2\), so the y-intercept is \((0, a r_1 r_2)\).

Detailed Explanation

Quadratic function: y-intercept formula — step-by-step explanation

  1. Definition of y-intercept

    The y-intercept of a graph is the point where the graph crosses the y-axis. On the y-axis the x-coordinate equals zero, so to find the y-intercept you evaluate the function at x = 0.

  2. Start from the standard form of a quadratic

    The general (standard) form of a quadratic function is

    \[ y = ax^{2} + bx + c \]

    Here a, b, c are constants and a is nonzero for a true quadratic.

  3. Evaluate at x = 0

    Substitute x = 0 into the expression to get the y-value at the y-axis:

    \[ y(0) = a\cdot 0^{2} + b\cdot 0 + c \]

    Compute each term:

    • \(0^{2} = 0\), so \(a\cdot 0^{2} = 0\).
    • \(b\cdot 0 = 0\).
    • The remaining term is \(c\).

    Therefore

    \[ y(0) = c \]

  4. Conclusion for the y-intercept

    The y-intercept point is

    \[ (0,\, c) \]

    So the y-intercept formula for a quadratic in standard form is simply the constant term c.

  5. Formulas when the quadratic is given in other common forms

    If the quadratic is given in vertex form

    \[ y = a(x – h)^{2} + k \]

    then evaluate at x = 0:

    \[ y(0) = a(0 – h)^{2} + k = a h^{2} + k \]

    so the y-intercept is

    \[ (0,\, a h^{2} + k) \]

    If the quadratic is given in factored form

    \[ y = a(x – r_{1})(x – r_{2}) \]

    then evaluate at x = 0:

    \[ y(0) = a(0 – r_{1})(0 – r_{2}) = a r_{1} r_{2} \]

    so the y-intercept is

    \[ (0,\, a r_{1} r_{2}) \]

  6. Special note

    If the constant term c is not explicitly given, compute it by evaluating the polynomial at x = 0 or convert the given form (vertex or factored) to standard form and read off c.

Summary: For a quadratic in standard form y = ax^{2} + bx + c the y-intercept is c and the point is (0, c). For vertex form y = a(x – h)^{2} + k the y-intercept is a h^{2} + k. For factored form y = a(x – r_{1})(x – r_{2}) the y-intercept is a r_{1} r_{2}.

See full solution

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FAQs

What is the y-intercept of a quadratic y = ax^2 + bx + c?

The y-intercept is c, because evaluating at x = 0 gives y(0) = a·0^2 + b·0 + c = c.

How do I find the y-intercept from vertex form y = a(x - h)^2 + k?

Set x = 0: y(0) = a(0 - h)^2 + k = a h^2 + k. So the y-intercept is a h^2 + k.

How do I get the y-intercept from factored form y = a(x - r1)(x - r2)?

Evaluate at x = 0: y(0) = a( - r1)( - r2) = a r1 r2, which equals c in standard form.

Is the y-intercept always a real number?

Yes for real-coefficient quadratics: plugging x = 0 yields the real number c. If coefficients are complex, the y-intercept may be complex.

How is the constant term c related to the roots r1 and r2?

For ax^2 + bx + c, the product of roots satisfies r1 r2 = c / a, so c = a r1 r2.

How does changing c affect the graph of y = ax^2 + bx + c?

How does changing c affect the graph of y = ax^2 + bx + c?

Can the y-intercept equal the vertex's y-coordinate?

Only if the vertex x-coordinate h = 0. In vertex form y = a(x - h)^2 + k, if h = 0 then y-intercept = k (the vertex's y).

How do I read the y-intercept from a graph of a parabola?

Find the point where the curve crosses the y-axis (x = 0); the y-coordinate of that point is the y-intercept. If it doesn't cross visibly, estimate or use the equation.
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