Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
4. Section 9.4
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Exercise 133 Page 531

Practice makes perfect
a Let's start with a(x). There's nothing that indicates that this function has the shape of a parabola or any other polynomial function of a higher degree. Therefore, we will assume it is either linear or exponential.
Linear:& y=mx+b Exponential:& y=ab^x Examining the table's three first data points, we notice that it cannot be linear as this would imply a constant rate of change, which is not the case here. If it is an exponential function, there is a common ratio between consecutive data points. Let's investigate this.
As we can see, we have a common ratio between consecutive data points which means this is in fact an exponential function with an initial value of a=10 and a common ratio of b=1.05. With this information, we can write the function. a(x)= 10(1.05)^x Next we will have a look at b(x). From the graph, we notice that it has the shape of a parabola which means this is a quadratic function. We also see that it has its vertex in (-2,24) and a y-intercept of (0,24). When we know the vertex of a parabola we can write it in graphing form. Graphing Form:& f(x)=a(x-h)^2+k Vertex:& (h,k) Let's substitute h=-2 and k=4 into the graphing form of a quadratic. b(x)=a(x-(-2))^2+4 ⇕ b(x)=a(x+2)^2+4 To find the value of a we will substitute the y-intercept into the function and then proceed to isolate a by performing inverse operations.
b(x)=a(x+2)^2+4
24=a( 0+2)^2+4
Solve for a
24=a(2)^2+4
24=a(4)+4
20=a(4)
5=a
a=5
Now we can complete the equation. b(x)=5(x+2)^2+4
b Let's start by sketching the functions on the same coordinate plane.

When x gets large, a(x) will at some point intersect b(x) and then exceed it.