Quadratic Functions 8-8
A QUADRATIC FUNCTION IS NOT y = mx + b
(which is a LINEAR function),
but instead is
y = ax2 + bx + c
OR
f(x) = ax2 + bx + c
where a, b, and c are all real numbers and
a cannot be equal to zero because
it must have a variable that is squared (degree of 2)
Quadratics have a squared term, so they have 2 possible solutions (roots)
You already saw this when you factored the trinomial and used zero products property.
If the domain is all real numbers, then you will have a PARABOLA which looks like a smile when the a coefficient is positive or
looks like a frown when the a coefficient is negative.
Let's look at f(x) = x2 - 2x - 2
When we plotted a few points, we discovered that definitely was NOT a straight line.
We used f(0), f(1), f(2), f(3), f(4), and then f(-1), f(-2)
we connect with a smooth curved line a U shape with arrows at either end--> because the parabola continues without end. Do not put the equation of the parabola on it, however.
Notice this parabola opens upward and has a minimum point or lowest point at (1, -3)
the y coordinate at this point is the least value of the function.
Domain: {x Ι x = R} This is read as "the Domain is x such that x equals all real numbers." and the
Range: { y Ι y ≥ -3} This is read as "the Range is y such that y is greater than or equal to -3."
the vertical line x = 1 contains this minimum point and is called the axis of symmetry. If you fold the graph along the axis of symmetry the two halves coincide. You could always just plot points-- but how would you know which points to pick?
Let's look at f(x) = -x2 + 2x + 2
Just by looking at this function I can tell that it opens downward. That (0,2) is where one side of the parabola crosses the y-intercept.
f(x) = ax2 + bx + c ( a ≠ 0 ) is a quadratic function. Why can't a =0?
If the domain of f is the set of all real numbers, then the graph is a parabola
if a in the function f(x) =ax2 + bx + c is positive, the graph opens upward. It's a happy face!
If a in the function f (x) = ax2 + bx + c is negative, the graph opens downward. It's a sad face!
The minimum or maximum point is called the vertex of the parabola.
Notice that all the other points ( except the vertex) occur in pairs with the same y-coordinate-- and the same distance away from the axis of symmetry. That makes sense, doesn't it?
The x coordinate of the vertex of a parabola f(x) = ax2 + bx + c
is -b/2a
YES... you do need to memorize this!
The axis of symmetry is the line x = -b/2a
H:x --> 2x2 + 4x -3
Find the vertex
Remember -b/2a is the x coordinate of the vertex
so
-b/2a = -4/2(2) = -1
The x coordinate of the vertex is -1. Plug in this x value to find the corresponding y value for the vertez
f(-1) = 2(-1)2 + 4(-1) -3 = 2 -4 -3 = -5
so the vertex is
(-1, -5)
and
x = -1 is the line of symmetry or the AXIS OF SYMMETRY
What happens as the "a" coefficient gets really big or really small (fraction/decimal)? We'll look at that together on my graphing calculator.
But think, what happened when the "m" (slope) coefficient got big?
The slope got steeper.
So now think that both sides of the U get steeper at the same time.
What's happening to the shape of the U???
Now think, what happened when the "m" (slope) coefficient got tiny?
The slope was a bunny slope.
So now think that both sides of the U are bunny slopes at the same time.
What's happening to the shape of the U???
Putting in standard form:
Standard form is:
y = ax2 + bx + c
OR
f(x) = ax2 + bx + c
You can't read the sign of a, b, or c until it's in standard form (just like y = mx + b!)
Graphing quadratics:
You can graph quadratics exactly the way you graphed lines ...by plugging in your choice of an x value and using the equation to find your y value.
Because it's a U shape, you should graph 5 points as follows:
First MAKE SURE THE EQUATION IS IN STANDARD FORM!
y must be isolated on one side and then you can read the a and b coefficients.
y = ax2 + bx + c
Point 1) the vertex - the minimum value of the smile or the maximum value of the frown
The x value of the VERTEX = -b/2a
Plug that into the equation and then find the y value of the vertex
Next, draw the AXIS OF SYMMETRY :
x = -b/2a
a line through the vertex parallel to the y axis
Point 2) Pick an x value IMMEDIATELY to the right or left of the AXIS OF SYMMETRY and find its
y by plugging into the equation.
Point 3) Graph its mirror image on the other side of the AXIS OF SYMMETRY by counting from the axis of symmetry
Points 4 and 5) Repeat point 2 and 3 directions with another point ONE STEP FARTHER from the AXIS OF SYMMETRY.
JOIN YOUR 5 POINTS IN A SMOOTH "U" SHAPE ( not a V shape!)
AND EXTEND LINES WITH ARROWS ON END
Parabolas are functions whose domains are ALL REAL NUMBERS.
Their ranges depend on where the vertex is and also if the ‘a’ coefficient is positive or negative
EXAMPLE: f(x) = -3x2
the ‘a’ coefficient is negative so it is a frowny face
The vertex is called the maximum.
The x value of the vertex is -b/2a
a = -3 and b = 0 (it's missing!)
The x value of the vertex = -b/2a = -0/2(-3) = 0
Plug the x value of 0 back into the function to find the y value of the
vertex:
y = -3(02) = 0 So the vertex is (0, 0)
The domain is all real numbers.
The range is y is less than or equal to zero (It's a frowny face)
To graph this function:
1) Graph vertex (0, 0)
2) Draw the AXIS OF SYMMETRY –
a dotted line at x = 0 (actually this is the y axis!)
3) Pick x value immediately to the right of axis of symmetry, x = 1
Plug it in the equation to find the y value: y = -3(1) = -3
Plot (1, -3)
4) Count the same 1 step from axis of symmetry on the other side of the axis and place another point to the LEFT of axis at the same y value (-1, -3)
5) Pick another x value to the right 2 steps away from the axis of symmetry, x = 2
Plug it in the equation to find y:
y = -3(22) = -12 Plot (2, -12)
6) Count 2 steps from axis of symmetry on the other side of it and place another point to the LEFT of axis at the same y value (-2, -12)
JOIN YOUR 5 POINTS IN A "U" SHAPE AND EXTEND LINES WITH ARROWS ON END
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