SOLUTION: Can you graph the equation: y= -X^2-4X+5. If you can, please explan how u got the vertex and how to graph everything. Thanks.

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Question 142145: Can you graph the equation: y= -X^2-4X+5. If you can, please explan how u got the vertex and how to graph everything. Thanks.
Answer by nabla(475) (Show Source):
You can put this solution on YOUR website!

First of all, let's find where the function intercepts the x-axis by setting f(x)=0.

<-- Make it a little easier by multiplying both sides by -1.
<---Factor-thought step: what numbers multiply to get -5 and add to get 4?
<--- 5 and -1!
Now, when can this equation be true? That is, if it equals 0, one of either term (x+5) or (x-1) must be 0. So we set each equal to zero.
x+5=0 or x-1=0
which means x=-5 or x=1. So we now have the x-intercepts.
Next, the vertex of a parabola is -b/2a, when the function is in form ax^2+bx+c. So, for our purposes the vertex is x=-(-4)/2(-1)=4/(-2)=-2. The function is valued at f(-2)=-(-2)^2-4(-2)+5=-4+8+5=9. So the vertex is (-2,9). Note the direction in which the function opens: the leading coefficient is -1 (of x^2). This means it opens downward.
Now you can graph it properly. Look at my graph below and see if you can reproduce it yourself.