I'm teaching myself math. I was working on $y = -x^2 + 8x$ and I used the quadratic equation to determine the x-coordinates. I ended up with $(0,0)$ and $(-8,0)$. The vertex is $(4,16)$ and the y-intercept is $(0,0)$. The thing is, x coordinate of root should be a positive number$=+8,$ otherwise I cannot have a parabola.
Where is my error?
On
The thing is, 8 should be a positive number, otherwise I cannot have a parabola.
This makes no sense. Also, all your calculated points are correct. It may relieve your confusion for you to know that the parabola is concave (i.e opens downward) and the vertex is above the curve and on the positive side of the y-axis.
On
Actually, you have the parabola $$y = (-x^2 + 8x -16) +16 \iff y= -(x^2-8x+16)+16 \iff (y-16)= -(x-4)^2$$
This parabola opens downward (we see the coefficient before the squared function of x is negative (-1).
A parabola given by $y-b = m(x-a)^2$ has vertex at $(a, b)$. In this case, that makes the vertex $(4, 16)$.
The parabola intersects the x-axis when $y= 0$, which means, in this case, when $x^2-8x = x(x-8) = 0.$ This means that either $x=0$, or $x-8 = 0 \iff x = 8$. So the points $(0, 0)$ and $(8, 0)$ are the two points of the parabola at which the parabola intersects the x-axis. It seems your work was fine, except for the sign error with respect to the point $(0, 8).$
On
$y = -x^2 + 8x ; \, 0= x( 8-x) , \rightarrow x= +8,0 $
It is a simple sign error you made. Maximum value is correct.
On
The first thing I notice from the equation $y=-x^2+8x$ is the leading negative sign. This means the parabola opens downward. You don't state that you know this fact. However, since the vertex you give is correct and lies above the $x$-axis, you should have realized this to be true.
So you decided the intercepts were at $(0, 0)$ and $(-8, 0)$. The $x-$coordinate of the vertex should lie between these two values (indeed, at the average). But you noted the vertex is at $(4, 16)$, not $(-4, 16)$.
As a former teacher, I can tell you that most students would accept these answers as true, and not check to see if they make sense. You have checked, and for that I congratulate you. To reward this good habit of yours, I have written this long(ish) post.
I think the mistake you made was in seeing the plus sign in the original equation, $y=-x^2+8x$. When we see $x+8$, we immediately recognize $x=-8$. However, to properly factor this equation, you need to factor out $-x$, not just $x$. Like this
$y=-x(x-8)$
...so now we see that $x=8$, and the intercept is $(8,0)$, as you correctly surmised it must be.
You have that $-x^2 + 8x = -x(x-8)$ so the intersections of the parabola with the $x$-axis are $(0,0)$ and $(8,0)$ as you remarked. You probably made a sign mistake somewhere along your computations.