Question

Which statements about the graph of the function f(x) = -x2 - 4x + 2 are true? Select three options.
The domain is {x\x 5-2}
The range is vy s 6)
The function is increasing over the interval (-0, -2).
The function is decreasing over the interval (-4, 0).
The function has a positive y-intercept

Answer 1

Answer:

Domain is all real numbers.

The range is $\{y|y\le 6\}$

The function is increasing over $(-\infty,-2)$.

The function is decreasing over $(-2,\infty)$

The function has a positive y-intercept.

-----------------This is a guess if I had interpreted your choices correctly:

Second option: The range is $\{y|y\le 6\}$

Third option:  The function is increasing over $(-\infty,-2)$.

Last option:  The function has a positive y-intercept.

I can't really read some of your choices. So you can read my above and determine which is false. If you have a question about any of what I said above please let me know.

Note: I guess those 0's are suppose to be infinities? I hopefully your function is $f(x)=-x^2-4x+2$.

Step-by-step explanation:

$f(x)=-x^2-4x+2$ is a polynomial function which mean it has domain of all real numbers.  All this sentence is really saying is that there exists a number for any value you input into $-x^2-4x+2$.

Now since the is a quadratic then it is a parabola. We know it is a quadratic because it is comparable to $ax^2+bx+c$, $a \neq 0$.

This means the graph sort of looks like a U or an upside down U.

It is U, when $a>0$.

It is upside down U, when $a<0$.

So here we have $a=-1$ so $a<0$ which means the parabola is an upside down U.  

Let's look at the range.  We know the vertex is either the highest point (if $a<0$) or the lowest point (if $a>0$).

The vertex here will be the highest point, again since $a<0$.

The vertex's x-coordinate can be found by evaluating $\frac{-b}{2a}$:

$\frac{-(-4)}{2(-1)}=\frac{4}{-2}=-2$.

So the y-coordinate can be found by evaluate $-x^2-4x+2$ for $x=-2$:

$-(-2)^2-4(-2)+2$

$-(4)+8+2$

$4+2$

$6$

So the highest y-coordinate is 6.  The range is therefore $(-\infty,6]$.

If you picture the upside down U in your mind and you know the graph is symmetrical about x=-2.

Then you know the parabola is increasing on $(-\infty,-2)$ and decreasing on $(-2,\infty)$.

So let's look at the intevals they have:

So on $(-\infty,-2)$ the function is increasing.

Looking on $(-4,\infty)$ the function is increasing on (-4,-2) but decreasing on the rest of that given interval.

The function's y-intercept can be found by putting 0 in for $x$:

$-(0)^2-4(0)+2$

$-0-0+2$

$0+2$

$2$

The y-intercept is positive since 2>0.