8a - Lab 8 Quadratic Functions.sagews

##Lab 8 Quadratic Functions In this lab you will review the solution processes for finding zeroes of quadratic functions. You will also examine different forms of a quadratic function. First, there are 3 solution processes you can use when solving a quadratic by hand: factoring, complete the square and the quadratic formula. Let's only look at the quadratic formula since it will work fastest.(Of course we can just tell SAGE to SOLVE it!)

#study my lines of code, then execute
a,b,c,x=var('a,b,c,x')
#I am going to create a quadratic function given its three coefficients: a, b and c
general_quad(a,b,c,x)=a*x^2+b*x+c
general_quad(1,2,3,x);  show(general_quad(1,2,3,x))#an example
general_quad(1,2,3,x).plot()
#I am going to create a "list" of the solutions as calculated by the quadratic formula.
solutions(a,b,c)=[(-b+sqrt(b^2-4*a*c))/(2*a),(-b-sqrt(b^2-4*a*c))/(2*a)]
solutions(1,2,3);  show(solutions(1,2,3)); #You'll notice that SAGE doesn't turn the complex solutions into the imaginary form
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Your turn. . . Use copy and paste with the code above to create, graph and find the solutions to these three:
1.  $3x^2-12x+10$
2.  $-\frac{x^2}{2}-4x-12$
3.  $x^2-19$

Your turn. . . Use copy and paste with the code above to create, graph and find the solutions to these three:

1. $3x^2-12x+10$

2. $-\frac{x^2}{2}-4x-12$

3. $x^2-19$

The above code created quadratics in the general form. Let's take a moment and examine the factored form. Here we'll create a quadratic in its factored form. You will see that the zeroes of this type will just be constants inside the binomial factors.

#examine code before executing
m,n=var('m,n')
factored_quad(a,m,n,x)=a*(x-m)*(x-n);  show(factored_quad(a,m,n,x))
factored_quad(1,3,-2);  show(factored_quad(1,3,-2))
#solving the factored quadratic after setting equal to zero
solutions_factored(a,m,n)=solve(factored_quad(a,m,n,x)==0,x)
solutions_factored(1,3,-2);  show(solutions_factored(1,3,-2))#Of course we already know that the zeros are m and n!
#m and n, being zeros, will also be the x-intercepts of the parabola(if they're real)
plot(factored_quad(1,3,-2,x),(x,-3,4))
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%md
Create factored form quadratics and graph them for the following criteria:
1.  leading term coefficient is 1/2.  Zeros are 5 and $\sqrt{2}$
2.  leading term coefficient is .03.  x-intercepts are $3\sqrt{3}$ and $-2\sqrt{3}$
3.  leading term coefficient is -3.  Zeros are both -4.

Create factored form quadratics and graph them for the following criteria:

1. leading term coefficient is 1/2. Zeros are 5 and $\sqrt{2}$

2. leading term coefficient is .03. x-intercepts are $3\sqrt{3}$ and $-2\sqrt{3}$

3. leading term coefficient is -3. Zeros are both -4.

The vertex form of a quadratic depends on the coordinates of the vertex of the resulting parabola and also the vertical line of the form $x=h$ called the axis of symmetry.

h,k=var('h,k')
vertex_quad(a,h,k,x)=a*(x-h)^2+k;  show(vertex_quad(a,h,k,x))
#Here's a parabola that opens up with a vertex = (2,3)
vertex_quad(.5,2,3,x);  show(vertex_quad(-5,2,3,x))
#Now to plot it along with the axis of symmetry.  You will notice a new plot command called 'line'
p1=plot(vertex_quad(.5,2,3,x),xmin=-10,xmax=10)#the parabola
p2=line([(2,-20),(2,70)],linestyle="--",color='red')#the axis of symmetry
vertex=point((2,3),size=30,color='black')#the vertex
p1+p2+vertex#plot the parabola, the axis of symmetry and the vertex
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%md
Use the above code to produce parabolas with these characteristics.  Graph the parabolas, alongwith the axis of symmetry and vertex point.
1.  Has a vertex at (-4,5) and opens downward.
2.  Has a vertex at (5,-6) and opens upward.
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One of the questions you will probably be asked is to create a quadratic in vertex form that has a particular vertex, (h,k), and also has a particular y-intercept.  Below I direct SAGE to create and graph a parabola that has its vertex at (5,8) and its y-intercept at (0,1).  I'll use a couple of code blocks since I need to calculate as I work.  the first thing I have to do is determine the "a"-value since I know (h,k).  I know that when x=0, y=1 so I will use this info to find a.

One of the questions you will probably be asked is to create a quadratic in vertex form that has a particular vertex, (h,k), and also has a particular y-intercept. Below I direct SAGE to create and graph a parabola that has its vertex at (5,8) and its y-intercept at (0,1). I'll use a couple of code blocks since I need to calculate as I work. the first thing I have to do is determine the "a"-value since I know (h,k). I know that when x=0, y=1 so I will use this info to find a.

a=var('a')
#I will replace x, h, k with 0, 5 and 8 and set equal to 1(the y-intercept)
a*(0-5)^2+8==1;  show(a*(0-5)^2+8==1)
#Now I will solve this equation for 'a'
solve(a*(0-5)^2+8==1,a);  show(solve(a*(0-5)^2+8==1,a))

25*a + 8 == 1

$\displaystyle 25 \, a + 8 = 1$

[a == (-7/25)]

[$\displaystyle a = \left(-\frac{7}{25}\right)$]

#Now to use this a-value to create the plot.  Being efficient(lazy), I'll copy and paste from previous work
p1=plot(-7/25*(x-5)^2+8,xmin=-10,xmax=10)#the parabola in the form, a*(x-h)^2+k
p2=line([(5,-20),(5,70)],linestyle="--",color='red')#the axis of symmetry
vertex=point((5,8),size=30,color='black')#the vertex
p1+p2+vertex#plot the parabola, the axis of symmetry and the vertex

Create and graph the parabola with vertex, (-1,-2), and y-intercept=6.

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