The curve y=6-x-2x^2.

Given the curve y=6-x-2x^2 determine the following key points if they exist. The x intercept, the y intercept, and the vertex. Just wondering could someone please help me identify and work out how to get the x intercept, y intercept, and the vertex. Thanx

The vertex is at dy/dx = 0

y = 6 - x - 2x^2
=> dy/dx = -4x - 1
=> 0 = -4x - 1
=> 4x = -1
=> x = -1/4

y = 6 - (-1/4) - 2 * (-1/4)^2
=> y = 6 + 1/4 - 2/16
=> y = 96/16 + 4/16 - 2/16
=> y = 6 1/8 aka 49/8

so the vertex is at (-1/4, 49/8)

the y-intercept is when x = 0

=> 6 - 0 - 2*0^2 = 6
=> y-intercept (0,6)

the x-intercept is when y = 0

=> 0 = 6 -x - 2x^2
=> 0 = 6 +3x - 4x - 2x^2
=> 0 = 3(2 + x) -2x(2 + x)
=> 0 = (2 + x)(3 - 2x)
=> x = -2 and 3/2

=> x- intercepts are (-2, 0) and (3/2, 0)

The Vertex is the turning point of the parabola, the point where the curve starts going the other way. it's the minimum point on an upwards opening parabola, and the maximum point on a downward opening parabola

Report Abuse

The y intercept:
Put x=0, and the value of y intercept is obtained.
=6
x- intercept =
The difference between the zeroes of the curve.
y= (3 -2x)(x+2)
so the zeroes are:
3/2 and -2
and the diff. = 3/2 - (-2)
= 7/2 = x intercept.
Now, the given equation is a quadratic which represents a downward opening parabola.
The vertex is given by the formula:
vertex = ( -b/2a , -D/4a)
where D = b² - 4ac