02.13quadgraph

Quadratic Graphs (Parabolas)

A second degree polynomial forms a quadratic graph (parabola)

  • {pronounced pa-RAB-ola}

General Form

The general form of the quadratic function is $y = ax^2 + bx + c$

The sign of a (the co-efficient of x2) controls the direction of the curve

  • a > 0, the graph has a minimum turning point, parabola is upright (with open end up)
  • a < 0, the graph has a maximum turning point, parabola is inverted (open end down)

The constant term, c, is the y-intercept

The turning point is at $x = -\dfrac{b}{2a}$
OR the turning point can be found by taking the derivative and setting it to zero

The parabola is symmetrical on each side of the turning point

The turning point is midway between the two x-intercepts (ie it is the average of the two x-intercepts)

02.13quadform.gif

The parabola may have two or one or no x-intercepts
The x-intercepts are given by the solution to $ax^2+ bx + c = 0$

  • this can be solved by factorising, then using the Null Factor Law
  • or by using the Quadratic Formula

The Discriminant

The Discriminant is the part inside the square root of the Quadratic Formula

  • we use the greek letter Delta, $\Delta$ for the Discriminant}

… … $\Delta = b^2 - 4ac$

  • If $\Delta > 0$, the graph will have 2 x-intercepts
  • If $\Delta = 0$, the graph will have 1 x-intercept (a turning point)
  • If $\Delta < 0$, the graph will have no x-intercepts

Example 1

Sketch $f: R \longrightarrow R, f(x) = 2x^2 + 3x - 5$

Solution:

y-intercept (let x = 0):

… … $y = -5$

x-intercepts:

… … $2x^2 + 3x - 5 = 0$

… … $(2x + 5)(x - 1) = 0$

… … $2x + 5 = 0 \quad \text{ OR } \quad x - 1 = 0$

… … $x = -2.5 \qquad \quad \text{ OR } \qquad \quad x = 1$

Turning point:

… … Because a > 0,

… … … … turning point is a minimum,

… … … … parabola is upright.

… … turning point is halfway between –2.5 and 1

… … $x = \dfrac{-2.5 + 1}{2}$

… … $x = - \frac{3}{4}$

… … $f \left( -\frac{3}{4} \right) = -6\frac{1}{8}$

… … so, coordinates of turning point are : $\; \Big( -\dfrac{3}{4}, \; -6\dfrac{1}{8} \Big)$

… … $\textbf{Domain: } \quad \; x \in R$

… … $\textbf{Range: } \qquad y \in \left[ -6\frac{1}{8}, \; \infty \right)$

02.13graph1.gif

Turning Point Form

The turning point form of the quadratic function is $y = a(x - h)^2 + k$

The usual transformations apply:

  • The sign of a controls the direction of the curve
    • a > 0, the graph has a minimum (open end up)
    • a < 0, the graph has a maximum (open end down)
    • The magnitude of a controls the dilation of the curve
      • a > 1, (or a < –1) the graph is thinner
      • a < 1, (or a > –1) the graph is wider
  • The turning point is at (x = h, y = k)
    • The standard parabola y = x2 has been shifted right by h and up by k
  • To find the y-intercept, substitute x = 0
  • To find the x-intercept, substitute y = 0

Example 2

… … Sketch $y = -\frac{1}{2} \big( x-3 \big)^2 +1$

Solution:

Turning Point

… … (3, 1) maximum (parabola is inverted)

Dilation

… … Wider (Dilation factor = 0.5)

Y-intercept

… … $y = -\frac{1}{2} \big( -3 \big)^2 +1$

… … $y = -3\frac{1}{2}$

X-intercepts

… … $-\frac{1}{2} \big( x-3 \big)^2 +1 = 0$

… … $-\frac{1}{2} \big( x-3 \big)^2 =-1$

… … $\big( x - 3 \big)^2 = 2$

… … $x-3=\pm \sqrt{2}$

… … $x=3\pm\sqrt{2}$

02.13graph2.gif

Changing into Turning Point Form

A quadratic in general form can be changed into turning point form by completing the square.

Example 3

… … Change $y = x^2 + 8x + 1$ into turning point form

Solution:

… … $y = x^2 + 8x + 1$

halve the middle term and square to get 16. Add 16 then subtract 16 again

… … $y = x^2 + 8x + 16 - 16 + 1$

the first three terms are now a perfect square, the last two terms will simplify

… … $y = (x + 4)^2 - 15$

This gives an upright parabola with

  • minimum turning point at (–4, –15),
  • dilation factor = 1.

Example 4

Change $y = -2x^2 + 12x - 14$ into turning point form by completing the square.

Solution:

Can't complete the square when x2 has a coefficient other than 1

… … $y = -2x^2 + 12x - 14$

Take out –2 as a common factor of the first two terms only

… … $y = -2(x^2 - 6x) - 14$

Halve the coefficient of x and square to get 9. Add 9 then subtract it again. Inside the brackets!

… … $y = -2(x^2 - 6x + 9 - 9) - 14$

the first three terms are now a perfect square.
Move the –9 out of the bracket by multiplying it by –2

… … $y = -2(x - 3)^2 + 18 - 14$

… … $y = -2(x - 3)^2 + 4$

This gives an inverted parabola with

  • maximum turning point at (3, 4),
  • dilation factor = 2.(so thinner)

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