01Equadraticgraphs

toc = 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 x 2 ) 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 = –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[[image:01Equad3.gif align="right" caption="Quadratic Formula"]]
 * The turning point is midway between the two x-intercepts (ie it is the average of the two x-intercepts)
 * 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, D for the Discriminant}

... ... D = b 2 – 4ac
 * If D > 0, the graph will have 2 x-intercepts
 * If D = 0, the graph will have 1 x-intercept (a turning point)
 * If D < 0, the graph will have no x-intercepts

Example 1


Sketch f: R → R, f(x) = 2x 2 + 3x – 5


 * Solution:**

y-intercept: ... ... y = –5

x-intercepts: ... ... 2x 2 + 3x – 5 = 0 ... ... (2x + 5)(x – 1) = 0 ... ... 2x + 5 = 0 .. ** OR ** .. x – 1 = 0 ... ... x = –2.5 ..... ** OR ** .... x = 1

Turning point: ... ... Because a > 0, ... ... ... ... turning point is a minimum, ... ... ... ... parabola is upright. ... ... turning point is halfway between –2.5 and 1

math \\ . \qquad x = \dfrac{-2.5 + 1}{2} \\. \\ . \qquad x = - \frac{3}{4} \\. \\ . \qquad f \left( -\frac{3}{4} \right) = -6\frac{1}{8} \\. \\ . \qquad \text{so, coordinates of turning point are :} \; \Big( -\dfrac{3}{4}, \; -6\dfrac{1}{8} \Big) math . math \\ . \qquad \textbf{Domain: } \quad \; x \in R \\. \\ . \qquad \textbf{Range: } \qquad y \in \left[ -6\frac{1}{8}, \; \infty \right) math

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 = x 2 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

Demonstration
To view an interactive demonstration of parabolas using turning point form, go here.

Example 2


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


 * Solution:**

Turning Point ... ... (3, 1) maximum (parabola is inverted)

Dilation ... ... Wider (Dilation factor = 0.5)

Y-intercept math \\ . \qquad y = -\frac{1}{2} \big( -3 \big)^2 +1 \\. \\ . \qquad y = -3\frac{1}{2} math

X-intercepts math \\ . \qquad -\frac{1}{2} \big( x-3 \big)^2 +1 = 0 \\. \\ . \qquad -\frac{1}{2} \big( x-3 \big)^2 =-1 \\. \\ math math \\ . \qquad \big( x - 3 \big)^2 = 2 \\. \\ . \qquad x-3=\pm \sqrt{2} \\ .\\ . \qquad x=3\pm\sqrt{2} math

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 subtact 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 x 2 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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