Examining the equation, we notice that it contains a squared binomial where one of the terms is x. This means it is a quadratic equation whose graph has the shape of a parabola. If we move the constant to the right-hand side, we can write the equation in graphing form.
Now we can identify the vertex and stretch factor of the quadratic.
y= 1(x-2)^2+(-5) [0.2em]
[-1em]
&Vertex: (2,-5)
&Stretch Factor: 1
Let's graph the vertex and the parabola's line of symmetry. This is a vertical line that runs through the vertex x-coordinate.
To graph the parabola, we need to identify more points that fall on the graph. By rewriting the right-hand side into standard form, we can identify the graph's y-intercept.
The function's constant is -1 which means it intercepts the y-axis at (0,-1). Let's add this to the coordinate plane.
We can find a third point by first measuring the horizontal distance between the y-intercept and the line of symmetry. Then, we will move the same distance to the right of the line of symmetry. Once we have three points, the parabola can be graphed accurately.
b
To graph the inequality, we should first treat it as an equation. This will allow us to graph the inequality's boundary line.
Inequality:& y≤(x+3)^3
Equation:& y=(x+3)^3
To graph this equation, we can use the general equation of a cubic function.
y= a(x-h)^3+k
[-1.2em]
&Horizontal Translation: h
&Vertical Translation: k
&Stretch Factor: a
As we can see, the constants and the coefficient controls different transformations of the parent function.
y=x^3
When we add to the input, we get a translation to the left. Therefore, to obtain the given function, we have to translate the parent function by 3 units to the left.
Finally, we must shade the correct side of the boundary line. To do that, we will test a point that is not on the boundary line. In this case, the easiest point we can choose is the origin.
Since the inequality holds true, we should shade the side of the boundary line that contains the origin.
c
The general equation of a hyperbola allows us to identify its locator point, as well as the horizontal and vertical asymptote.
y=a(1/x-h)+k [0.2em]
[-1em]
&Locator Point: (h,k)
&Horizontal Asymptote: y=k
&Vertical Asymptote: x=h
Let's rewrite the function until it matches the general equation.
Having rewritten the function, we can identify the graph's locator point and asymptotes.
y=1(1/x-3)+4 [0.2em]
[-1em]
&Locator Point: (3,4)
&Horizontal Asymptote: y=4
&Vertical Asymptote: x=3
Now we can draw the asymptotes and locator point.
Next, we have to calculate some points that fall on the graph. We can do that with a value table. Notice that we cannot pick x=3 for input as this would make the denominator equal to 0.
|c|c|c|
[-0.8em]
x & 4+1/x-3 & y [0.8em]
[-0.8em]
1 & 4+1/1-3 & 3.5 [0.8em]
[-0.8em]
2 & 4+1/2-3 & 3 [0.8em]
[-0.8em]
2.5 & 4+1/2.5-3 & 2 [0.8em]
[-0.8em]
3.5 & 4+1/3.5-3 & 6 [0.8em]
[-0.8em]
4 & 4+1/4-3 & 5 [0.8em]
[-0.8em]
5 & 4+1/5-3 & 4.5 [0.8em]
By plotting the points in a coordinate plane, we can then draw the hyperbola.
