Analysing the Quadratic Function

Analysing the Quadratic Function — Moosa Academy

Given $f(x) = ax^{2} + bx + c$ , the three coefficients are enough to describe the curve completely — which way it opens, where its turning point lies, where it crosses each axis. This lesson works through that analysis as a fixed procedure.

Concept What each coefficient controls

a — opening direction and width. Positive opens upward, negative downward.

b — together with

a

, places the axis of symmetry.

c — the y-intercept, at the point

(0,\ c)

A larger $|a|$ makes the parabola narrower, a smaller one flattens it. The sign of $c$ simply lifts or lowers the whole curve.

Note The seven-step analysis

Write the function in standard form and read off $a$ , $b$ , $c$ .
Check the sign of $a$ to fix the opening direction.
Find the axis of symmetry, $x = -b/2a$ .
Substitute to find the vertex height.
Compute the discriminant $\Delta = b^{2} - 4ac$ .
Find the roots by factorising or by the quadratic formula.
Check that the results agree with the shape you predicted.

Example Finding the vertex

For $f(x) = x^{2} - 6x + 8$ , find the axis of symmetry and the vertex.

a = 1

b = -6

c = 8

x = -\dfrac{-6}{2(1)} = 3

f(3) = 9 - 18 + 8 = -1

⟹ axis x = 3, vertex (3, −1)

Example Roots by factorising

Find the roots of $f(x) = x^{2} - 5x + 6$ .

Set the function to zero:

x^{2} - 5x + 6 = 0

Look for two numbers with product 6 and sum −5: they are −2 and −3.

(x - 2)(x - 3) = 0

Each factor gives a root.

⟹ x = 2 and x = 3

The two roots average to 2.5, which is exactly the axis of symmetry — a useful check, since the roots are always placed symmetrically about it.

Example Using the quadratic formula

Solve $2x^{2} + 3x - 5 = 0$ .

a = 2

b = 3

c = -5

\Delta = 3^{2} - 4(2)(-5) = 9 + 40 = 49

\Delta > 0

and it is a perfect square, so the roots are rational.

x = \dfrac{-3 \pm 7}{4}

⟹ x = 1 and x = −2.5

Example A thrown ball

A ball is thrown upward from a height of 5 metres at 20 m/s. Its height is $h(t) = -5t^{2} + 20t + 5$ . Find the greatest height and when it occurs.

a = -5

b = 20

c = 5

. Since

a < 0

, the vertex is a maximum.

t = -\dfrac{20}{2(-5)} = 2

seconds

h(2) = -20 + 40 + 5 = 25

⟹ 25 metres, reached after 2 seconds

The coefficient $c = 5$ is the launch height, and $a = -5$ carries the effect of gravity — the physics reads directly off the coefficients.

Summary

a fixes the opening direction and width, b positions the axis, c is the y-intercept.
The vertex is at x = −b/2a, and its height comes from substituting back.
The discriminant b² − 4ac tells you the number and type of roots before solving.
The roots sit symmetrically about the axis, so their average must equal −b/2a.