c What is the y-intercept in the teacher's graph? What is your y-intercept?
A
aQuadratic Equation: y=x^2-4x+4
B
bCubic Equation: y=(x+2)(x^2-4x+4)
Graph:
C
cFunction: y= 14(x+2)(x^2-4x+4)
Graph:
Practice makes perfect
a We are told that the graph resembles a parabola with vertex (2,0) near x=2. Let's use the vertex form of a parabola to find the equation.
y = (x-h)^2+kBy substituting h= 2 and k= 0 we get the equation of the parabola. We will expand the perfect square to find its standard form.
y = (x- 2)^2
⇒
y = x^2 - 4x + 4
Let's make a plot of this parabola to check that we found the correct equation.
y = mx + b
In the equation above, m is the slope and b the y-intercept. We are asked to find the equation of a line with slope of 1 that passes through (-2,0). Then, let's substitute m= 1, x= -2, and y= 0 to find the value of b.
Therefore, the equation of the required line is y=x+2. Our next step is to multiply this equation by the quadratic equation we found in Part A.
y = (x+2)(x^2-4x+4)
Let's make the graph of the equation we just built.
By comparing the graph above with the one our friend's teacher showed, we can see that the zeros are the same but the y-intercept is not.
c In the previous part, we noticed that the y-intercept in the teacher's graph is 2 while our y-intercept is 8. We will multiply our equation by a constant k so that the y-intercept changes to 2.
y= k((x+2)(x^2-4x+4))
Since we want the y-intercept to be 2, we will substitute x= 0 and y= 2 and solve the resulting equation for k.
Consequently, our new equation is the one shown below.
y = 1/4(x+2)(x^2-4x+4)
Finally, let's make a plot of this equation to check that it looks like the one the teacher showed.
