02.15quarticgraphs

# Quartic Graphs A quartic function has degree 4.

The general form of a quartic is \$y = ax^4 +bx^3 +cx^2 + dx + e\$

The sign of a (the coefficient of x4) controls the direction of the curve

• a > 0, the graph starts in 2nd Quadrant and ends in 1st Quadrant (upright)
• a < 0, the graph starts in 3rd Quadrant and ends in 4th Quadrant (inverted)

The constant term (e) gives the y-intercept

A quartic may have zero, one, two, three or four x-intercepts

## The Basic Form (Power Form)The Basic Quartic

The basic quartic . \$y = x^4\$ . looks a bit like a parabola but has a wider base and steeper sides. The basic quartic can be dilated and shifted in the same way as other curves we have studied, producing the power form of the quartic:

… … \$y = a(x - h)^4 + k\$

The usual transformations apply:

• This quartic has a turning point at (h, k)
• And is dilated by a factor of a

## Factor Form

If a quartic can be factorised,

• Any linear factors will give the x-intercepts
• a factor repeated twice indicates a turning point on the x-axis at that x-value
• a factor repeated three times indicates a stationary point of inflection on the x-axis at that x-value
• a factor repeated four times indicates a turning point on the x-axis at that x-value
• (the graph is the basic power quartic shifted sideways) #### Example 1

Find the equation of the graph shown here. Solution

The graph is clearly a quartic.

By observation, the x-intercepts are x = –2, 1, 3

The turning point at –2 indicates a repeated factor

Hence the equation will be:
… … \$y = a(x - 1)(x - 3)(x + 2)^2\$

… … where a is a constant

Given that the quartic is inverted, we expect a to be negative.

The y-intercept is (0, –3) so substitute (0, –3) into the equation and solve for a

… … \$a(-1)(-3)(2)2 = -3\$

… … \$12a = -3\$

… … \$a = -\dfrac{1}{4}\$

Hence the equation is:

… … \$y = -\dfrac{1}{4}(x - 1)(x - 3)(x + 2)^2\$

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