Question

The funcilon f(x) = -(x + 5)(x +

Answer 1

Answer:

all numbers less than or equal to $-3$.

Step-by-step explanation:

The original equation shown is in "root form", a form of a quadratic equation showing where it hits the $x$-axis. This form does not show, like a "standard form" quadratic equation does, the vertex of the quadratic. The standard form equation is shown below.

$ax^2+bx+c$

1. $a$ is a constant showing the vertical stretch or compression of the quadratic. Also it can show if the function is flipped based on if it has a negative symbol at the front.
2. $b$ isn't very important when graphing unless you want to convert the equation to root-form.
3. $c$ is the height of the vertex. This is very important to know for this question because the vertex is either the minimum or maximum of the quadratic.

If we want to solve this equation showing all our work, the best bet is converting our equation from root-form to standard-form, like so:

$f(x)=-(x+5)(x+1)$

• distribute $x$ and $5$ into $(x+1)$. Then take this and distribute the negative sign.

$f(x)=-x^2-5x-x-5\\f(x)=-x^2-6x-5$

• This is our standard-form of the quadratic. Now we just need to determine the range of $y$ values it has.
• Note: The answers are wrong based on the $c$ value I have for this equation, so I'm going to just give the answer closest to this one.

Based on the properties of a standard form equation I gave earlier, this equation will look like an upside down "U" with a maximum value of $-5$. Therefore (based on the italicized text above), the range of $y$ values for this equation must be "all real numbers less than or equal to $-3$".

P.S: Message me with any questions or fixes I should make to my math.