Standard Form of a Quadratic

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Standard Form of a Quadratic: meaning, method and solved example (Algebra 1) – Maths PDF

In short: Standard form of a quadratic is the arrangement y = ax² + bx + c with a ≠ 0, where the three numbers a, b and c shape the curve.

Algebra 1Functions and GraphsF-IF.C.8a

What is Standard Form of a Quadratic?

Writing terms from the highest power downward gives the standard form of a quadratic, \(y = ax^2 + bx + c\), and each coefficient sends a message about the curve.

The number \(a\) gives the direction (up when positive, down when negative) and the width. The number \(c\) is the y-intercept, because \(x = 0\) erases the other terms. The pair \(a\) and \(b\) fix the mirror line at \(x = -\frac{b}{2a}\), and setting \(y = 0\) leads to the zeros.

How to spot it

  • The standard form of a quadratic lists the squared term first, then an x term, then a plain number.
  • A question supplies a, b and c and asks for direction, intercept or turning point.
  • Factoring or the quadratic formula is needed, and both begin from this arrangement set to zero.
  • Two binomials have been multiplied out and the result needs interpreting.

The method

  1. Arrange the powers in descending order and read a, b and c along with their signs.
  2. State the direction from the sign of a and the y-intercept from c.
  3. Compute \(x = -\frac{b}{2a}\) and substitute it to get the turning point.
  4. Solve \(ax^2 + bx + c = 0\) by factoring or by the quadratic formula for the x-intercepts.
(0, 6) = c(1, 0)(3, 0)vertex (2, −2)y = 2x² − 8x + 6a = 2, b = −8, c = 6axis: x = 8/4 = 2
Coefficients a = 2, b = −8, c = 6 give y-intercept 6 and axis x = 2.

Standard Form of a Quadratic: solved example

Problem

For \(g(x) = -x^2 + 4x + 12\), name a, b and c, then give the direction, the y-intercept, the mirror line and the turning point.

Solution

The coefficients are \(a = -1\), \(b = 4\), \(c = 12\). A negative a means the curve opens down, and the y-intercept is \((0, 12)\).

The mirror line is \(x = -\frac{4}{2(-1)} = 2\), and \(g(2) = -4 + 8 + 12 = 16\). The turning point is \((2, 16)\), a maximum.

Try it

Question 1

Name a, b and c in \(3x^2 – 5x + 2\).

Show the answer

\(a = 3\), \(b = -5\), \(c = 2\).

Question 2

Where does \(y = x^2 + 9x – 14\) cross the y-axis?

Show the answer

At the value of c, so \((0, -14)\).

Standard Form of a Quadratic vs. Vertex Form

The constant c here gives the y-intercept instantly, while the turning point needs a short calculation. The other layout flips that trade. Both describe a single curve, and expanding or completing the square links one to the other.

Vertex Form

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