For each of the functions below, state the domain and range, the restrictions, the intervals of increasing and decreasing, the roots, y-intercepts, and vertices.
1) f(x)=2x^2−8
2) f(x)=+√x - 2
3) f(x)=x+1/x-1
+118608 ok I will take a look :))
For each of the functions below, state the domain and range, the restrictions, the intervals of increasing and decreasing, the roots, y-intercepts, and vertices.
1) f(x)=2x^2−8
2) f(x)=+√x - 2
3) f(x)=x+1/x-1
$$1) \;\;f(x)=2x^2-8$$
Any real number can be squared so x can be any real number
Domain: $$x\in R$$ (x is in the set of reals) $$(-\infty,\infty)$$
$$x^2$$ cannot be negative, it can be any positive number though. Same for $$2x^2$$
$$2x^2-8$$ cannot be any smaller than -8
Range: $$-8\le f(x) < \infty\qquad \qquad [-8,\infty)$$
Even before you went to all this bother you should have recognised that this is a concave up parabola.
It is a steeper version of y=x^2 And every point is dropped 8 units.
In other words, The axis of symmetry is x=0 (That is the y axis)
The vertex is (0,-8)
The y intercept is -8
It is decreasing when x
It is increasing when x>0 (the tangent has a positive gradient)
When f(x)=0
$$\\ 0=2x^2-8\\
0=x^2-4\\
0=(x-2)(x+2)\\
x-2=0\qquad or \qquad x+2=0\\
x=2\qquad \qquad or \qquad x=-2\\
$The roots are x=2 and x=-2$$$
I think that covers the first one.
Think about it and aske questions if you have any.
Have a go at the second one yourself and present what you think.
Here is the graph:
https://www.desmos.com/calculator/gmxuxxwtbv
I will talk about it with you then :)
+118608 ok I will take a look :))
For each of the functions below, state the domain and range, the restrictions, the intervals of increasing and decreasing, the roots, y-intercepts, and vertices.
1) f(x)=2x^2−8
2) f(x)=+√x - 2
3) f(x)=x+1/x-1
$$1) \;\;f(x)=2x^2-8$$
Any real number can be squared so x can be any real number
Domain: $$x\in R$$ (x is in the set of reals) $$(-\infty,\infty)$$
$$x^2$$ cannot be negative, it can be any positive number though. Same for $$2x^2$$
$$2x^2-8$$ cannot be any smaller than -8
Range: $$-8\le f(x) < \infty\qquad \qquad [-8,\infty)$$
Even before you went to all this bother you should have recognised that this is a concave up parabola.
It is a steeper version of y=x^2 And every point is dropped 8 units.
In other words, The axis of symmetry is x=0 (That is the y axis)
The vertex is (0,-8)
The y intercept is -8
It is decreasing when x
It is increasing when x>0 (the tangent has a positive gradient)
When f(x)=0
$$\\ 0=2x^2-8\\
0=x^2-4\\
0=(x-2)(x+2)\\
x-2=0\qquad or \qquad x+2=0\\
x=2\qquad \qquad or \qquad x=-2\\
$The roots are x=2 and x=-2$$$
I think that covers the first one.
Think about it and aske questions if you have any.
Have a go at the second one yourself and present what you think.
Here is the graph:
https://www.desmos.com/calculator/gmxuxxwtbv
I will talk about it with you then :)
