You typed no equal sign.
If there is to be a vertex, it should be a conic, like a parabola. That means one of the variables must be squared.
How would I change this problem from standard form to vertex form?
The problem is:
y+6x-16
8 answers
Sorry. The equation is y=x+6x-16.
Could you possibly mean
y = (x+6)(x-16) ???
y = (x+6)(x-16) ???
No, that's not what I mean. I am supposed to be changing the equation y= x+6x-16 from standard form to vertex form.
That would lead you to
y = x^2 -10 x - 96
which is indeed a parabola. Since y gets big positive when x gets big positive or negative, it opens up (holds water)
Now complete the square to find the vertex and all
x^2 - 10 x = y + 96
add half of ten squared to both sides, in other words 25
x^2 - 10 x + 25 = y + 121
(x-5)^2 = y+121
vertex at (5 , -121)
y = x^2 -10 x - 96
which is indeed a parabola. Since y gets big positive when x gets big positive or negative, it opens up (holds water)
Now complete the square to find the vertex and all
x^2 - 10 x = y + 96
add half of ten squared to both sides, in other words 25
x^2 - 10 x + 25 = y + 121
(x-5)^2 = y+121
vertex at (5 , -121)
well
y= x+6x-16
is a straight line (a line has no vertex)
y = 7 x - 16
in slope intercept form
in standard form like Ax + By = C
that is
7 x - y = 16
y= x+6x-16
is a straight line (a line has no vertex)
y = 7 x - 16
in slope intercept form
in standard form like Ax + By = C
that is
7 x - y = 16
Could you mean
y = x^2 + 6 x - 16 ???
that would be
y+16 = x^2 + 6 x
add (6/2)^2 or 9 to both sides
y+25 = x^2 + 6 x + 9
y+25 = (x+3)^2
vertex at ( -3, -25)
y = x^2 + 6 x - 16 ???
that would be
y+16 = x^2 + 6 x
add (6/2)^2 or 9 to both sides
y+25 = x^2 + 6 x + 9
y+25 = (x+3)^2
vertex at ( -3, -25)
i need help with fractions
1 answer