I need to know the vertex of a equation?

Question

(x+1)(x-.67)=0
x^2-.67+7x-.67=0
x^2+.33x-.67=0

Answer 1

Find the vertex of the parabola.

y = (x + 1)(x - .67) = 0
y = x² + .33x - .67
y + (.33/2)² = (x² + .33x + (.33/2)²) - .67
y + 0.67 + (.33/2)² = (x² + .33x + (.33/2)²) - .67

y + .697225 = (x + .165)²

The vertex of the parabola is (-.165, -.697225).

Answer 2

I think you are asking about vertex of the quadratic function (expression on the left side of your equations). 'Vertex of an equation' does not make sense since a quadratic equation simply has solutions (two at most!), that's it. Quadratic function, however, does have a vertex. If you draw the quadratic function you will see a curve called parabola. If the parabola points up then the vertex is the x-coordinate of the lowest point on the parabola, and if it points down then the vertext is the x-coordinate of the highest point on the parabola.

Either way, if the function has a form of f(x) = a*x^2+b*x+c then the vertex is always calculated as -b/(2*a). Note that c doesn't matter since adding a constant to a function just shifts its graph and down, which clearly does not affect the vertex.

So for your example, f(x) = x^2+.33x-.67, so a = 1 and b = .33. Thus the vertex is -.33/2=-0.165.

Answer: -0.165

Answer 3

In the form ax^2+bx+c=y the vertex's x coordinate will be b^2/2a, then the equation can be solved with that x value for the y value of the vertex.


But the way you have given the equations, the solution will be a single point or two points.... It will not have a vertex

Answer 4

y = (x+1)(x-.67)
By symmetry, the vertex coordinates (h, k) can be found by
h = (-1+.67)/2 = -.165
k = ( -.165+1)( -.165-.67) = -.697225