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How do you find the vertex of an equation in standard form?
To find the vertex of a quadratic equation in standard form, (y = ax^2 + bx + c), you can use the vertex formula. The x-coordinate of the vertex is given by (x = -\frac{b}{2a}). Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. The vertex is then the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))).
How do you find the vertexof a parabola?
To find the vertex of a parabola given its equation in standard form (y = ax^2 + bx + c), you can use the formula for the x-coordinate of the vertex: (x = -\frac{b}{2a}). Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. Thus, the vertex can be expressed as the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))). For parabolas in vertex form (y = a(x-h)^2 + k), the vertex is simply the point ((h, k)).
How do you find the y-coordinate vertex of a parabola?
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.
Do you have find the croodinates for the vertex for the quadratic function?
Yes, the coordinates for the vertex of a quadratic function in the form (y = ax^2 + bx + c) can be found using the formula (x = -\frac{b}{2a}) to determine the x-coordinate. Once you have the x-coordinate, substitute it back into the original equation to find the corresponding y-coordinate. This gives you the vertex in the form ((x, y)).
How do you find an equation of a parabola using the vertex and a point it passes through?
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.
How do you find the vertex of an equation in standard form?
To find the vertex of a quadratic equation in standard form, (y = ax^2 + bx + c), you can use the vertex formula. The x-coordinate of the vertex is given by (x = -\frac{b}{2a}). Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. The vertex is then the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))).
How do you find the y-coordinate vertex of a parabola?
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.
The vertex form of the equation of a parabola is y x-5 2 plus 16 what is the standard form of the equation?
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.
How do you find the vertex from a quadratic equation in standard form?
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)
How do you find an equation of a parabola using the vertex and a point it passes through?
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.
How do you find the vertex of the parabola associated with a quadratic equation?
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.
What is the vertex of the graph -2x2 16x -15?
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).
What is the vertex of 2x2 plus 4x?
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).
How are the vertices of the parabolas related to the equation of the quadratic function?
Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a ≠0. The roots of the parabola are given by x = [-b ± 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 vertex of the parabola below is at the point (-4-2) which equation below could be one for parabola?
-2
Find equation what parabola its vertex is 0 0 and it passes through point 2 12 express the equation in standard form?
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
What is the x-coordinate of the of the vertex of y equals -3x2 plus 12x-5?
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).
