Q: What is the vertex of the parabola whose equation is y(x 1)2 3?

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"y = 2x2 - 12x + 6" is a quadratic equation which describes a parabola whose vertex occurs at the point (3, -12) and which has a range of -12 → ∞. It intercepts the x-axis at the points (3 - √6, 0) and (3 + √6, 0).

y=4x-12-3 is the equation of a straight line. It does not have a vertex. Did you mean y=x squared - 12x - 3 ?

y2-12=5x is an equation. When graphed, it is a parabola.

The x-intercept of an equation is any location where on the equation where x=0. In the case of a parabolic function, the easiest way to obtain the x intercept is to change the equation into binomial form (x+a)(x-b) form. Then by setting each of those binomials equal to zero, you can determine the x-intercepts.

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Y=3x^2 and this is in standard form. The vertex form of a prabola is y= a(x-h)2+k The vertex is at (0,0) so we have y=a(x)^2 it goes throug (2,12) so 12=a(2^2)=4a and a=3. Now the parabola is y=3x^2. Check this: It has vertex at (0,0) and the point (2,12) is on the parabola since 12=3x2^2

"y = 2x2 - 12x + 6" is a quadratic equation which describes a parabola whose vertex occurs at the point (3, -12) and which has a range of -12 → ∞. It intercepts the x-axis at the points (3 - √6, 0) and (3 + √6, 0).

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Without an equality sign in the given terms they can't be considered to be a straight line equation.

There are two forms in which a quadratic equation can be written: general form, which is ax2 + bx + c, and standard form, which is a(x - q)2 + p. In standard form, the vertex is (q, p). So to find the vertex, simply convert general form into standard form.The formula often used to convert between these two forms is:ax2 + bx + c = a(x + b/2a)2 + c - b2/4aSubstitute the variables:-2x2 + 12x - 13 = -2(x + 12/-4)2 -13 + 122/-8-2x2 + 12x - 13 = -2(x - 3)2 + 5Since the co-ordinates of the vertex are equal to (q, p), the vertex of the parabola defined by the equation y = -2x2 + 12x - 13 is located at point (3, 5)

You can calculate this by taking the derivative of the equation with respect to x, and solving it for a value of zero: y = x2 - 2x - 5 ∴ dy/dx = 2x - 2 Let dy/dx =0: 0 = 2x - 2 ∴ 2x = 2 ∴ x = 1 Now you can take that x value and plug it into the original equation to find it's y coordinate: y = 12 - 2(1) - 5 y = 1 - 2 - 5 y = -6 So the vertex of this parabola occurs at the point (1, -6).

y=4x-12-3 is the equation of a straight line. It does not have a vertex. Did you mean y=x squared - 12x - 3 ?

-2, 6

I think you are talking about the x-intercepts. You can find the zeros of the equation of the parabola y=ax2 +bx+c by setting y equal to 0 and finding the corresponding x values. These will be the "roots" of the parabola.

y2-12=5x is an equation. When graphed, it is a parabola.

x2 + 12x = 0 is an equation that describes a parabola. This parabola would have a minimum value and no maximum. That minimum can be found by taking its derivative and solving for zero: y = x2 + 12x dy/dx = x + 12 0 = x + 12 x = -12 Then take that x value and plug it in to the original equation: y = (-12)2 + 12(-12) y = 144 - 144 y = 0 So the focal point of the parabola is at (-12, 0) If you want to factor it, that would come to: x(x + 12) = 0

You can find the x-coordinate of it's vertex by taking it's derivative and solving for zero: y = -3x2 + 12x - 5 y' = -6x + 12 0 = -6x + 12 6x = 12 x = 2 Now that we have it's x coordinate, we can plug it back into the original equation to find it's y coordinate: y = -3x2 + 12x - 5 y = -3(2)2 + 12(2) + 5 y = -12 + 24 + 5 y = 17 So the vertex of the parabola y = -3x2 + 12x - 5 occurs at the point (2, 17).