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Q: How do you find the x coordinate of the vertex when we have an equation in standard form?

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Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.

I would use the alternate parabolic equation (y-k)2 = a (x-h). You can plug in the coordinates of the vertex (h, k)--h is the x coordinate of the vertex and k is the y coordinate of the vertex. Now you are left with an equation with an x and y, which are fine, but also an a, which we still have to get rid of. The a describes how steeply the parabola increases. To find the actual number, plug the other point you were given into the equation (where the x and y are). How you are left with one equation and one variable, so you can solve for a = some number. Now return to the beginning of the previous step, when you had and x, y, and a in your equation. Keep the x and y in their original positions, but replace a with the number you just found.

Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a â‰ 0. The roots of the parabola are given by x = [-b Â± sqrt(D)]/2a where D is the discriminant. Rather than solve explicitly for the coordinates of the vertex, note that the vertical line through the vertex is an axis of symmetry for the parabola. The two roots are symmetrical about x = -b/2a so, whatever the value of D and whether or not the parabola has real roots, the x coordinate of the vertex is -b/2a. It is simplest to substitute this value for x in the equation of the parabola to find the y-coordinate of the vertex, which is c - b2/2a.

The given equation is y = x - 4x + 2 which can be written as y = -3x + 2 This is an equation of a straight line. Therefore it has no vertex and so cannot be written in vertex form.

By solving the simultaneous equations the values of x and y should be equal to the given coordinate

Related questions

Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.

In the equation y x-5 2 plus 16 the standard form of the equation is 13. You find the answer to this by finding the value of X.

look for the interceptions add these and divide it by 2 (that's the x vertex) for the yvertex you just have to fill in the x(vertex) however you can also use the formula -(b/2a)

I would use the alternate parabolic equation (y-k)2 = a (x-h). You can plug in the coordinates of the vertex (h, k)--h is the x coordinate of the vertex and k is the y coordinate of the vertex. Now you are left with an equation with an x and y, which are fine, but also an a, which we still have to get rid of. The a describes how steeply the parabola increases. To find the actual number, plug the other point you were given into the equation (where the x and y are). How you are left with one equation and one variable, so you can solve for a = some number. Now return to the beginning of the previous step, when you had and x, y, and a in your equation. Keep the x and y in their original positions, but replace a with the number you just found.

You must first solve the equation for x, find the midpoint between the two x coordinates and substitute it into the equation to find the y coordinate I.E. If your equation was y=x^2+4x-5 1. Factorise the Equation and solve for x. x^2+4x-5 = (x+5)(x-1) So x+5=0 x = 0 - 5 x=-5 And x-1=0 x = 0 + 1 x = 1 So x = -5,1 Find the mid point/average: (-5+1)/2 = -2 So the x coordinate of the vertex is -2. Substitute this into the equation x^2+4x-5: (-2)^2 + 4*(-2) - 5 = 4 - 8 - 5 = -9 So the Coordinate of the vertex is (-2,-9). Hope this Helps, message me if you want more info.

Considering a general quadratic equation y=ax^2 + bx + c, the x coordinate of the vertex is found from the formula x= -b/2a and the y coordinate is found from putting that x coordinate back into the original quadratic equation which in this case I am assuming is y= -2x^2 + 16x -15. So, the x coordinate of the vertex is x=-16/(2*-2) = 4 To find the y coordinate we plug 4 back into y= -2x^2 + 16x -15 so we have y= -2 * 4^2 + 16*4 - 15. Following the order of operations we get y= -2 *16 + 64 - 15= 17 Therefore the vertex is at (4, 17).

We know that the x-coordinate of the vertex is x = -b/2a. We identify a, b, and c in f(x) = ax^2 + bx + c. y = 2x^2 + 4x or y = 2x^2 + 4x + 0 a = 2, b = 4, c = 0 Substitute 2 for a and 4 for b into the x-coordinate equation: x = -b/2a = -4/2(2) = -4/4 = -1 So the x-coordinate of the vertex is -1. To find the y-coordinate, substitute -1 for x in the equation of the function, y = 2x^2 + 4x. y = 2x^2 + 4x = 2(-1)^2 + 4(-1) = 2 - 4 = -2. Thus, the vertex is (-1, -2).

Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a â‰ 0. The roots of the parabola are given by x = [-b Â± sqrt(D)]/2a where D is the discriminant. Rather than solve explicitly for the coordinates of the vertex, note that the vertical line through the vertex is an axis of symmetry for the parabola. The two roots are symmetrical about x = -b/2a so, whatever the value of D and whether or not the parabola has real roots, the x coordinate of the vertex is -b/2a. It is simplest to substitute this value for x in the equation of the parabola to find the y-coordinate of the vertex, which is c - b2/2a.

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).

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

One way to solve this is to take the derivative of the equation with respect to x, and solve for zero. That will give you the x-coordinate of the vertex, which you can then plug back in to find its y-coordinate: y = x2 - 11x + 28 dy/dx = 2x - 11 0 = 2x - 11 x = 11/2 Now we plug it back in for the y-coordinate: y = (11/2)2 - 11(11/2) + 28 y = 121/4 - 121/2 + 28 y = -121/4 + 112/4 y = -9/4 So the vertex of the equation lies at (11/2, -9/4).

I'm assuming that you meant y = 2(x^2) +4. If it were only y = 2x +4, then this would be a linear equation and not a parabola. Anyways, use the equation x = -b/2a to find the x-value of your vertex AND your axis of symmetry. (Given the standard equation y = a(x^2) + bx + c) So, x = -0/2(2) - x = 0 (Axis of Symmetry) Now plug 0 back into your equation to find your y-value of your vertex. y = 2(0^2)+4 y=0 + 4 y = 4 Therefore Vertex = (0,4)