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Example:
Vertical velocity component of a 100 m/s cannonball fired at 30 degrees from horizontal = sin 30 * 100 = 50 m/s
Find the maximum height reached.
Take acceleration due to gravity at 10 (m/s)/s
so:
u = 50 m/s
v = 0
a = - 10 m/s/s (approx)
s = ?
Using:
v2 = u2 + ( 2 * a * s )
so:
s = ( v2 - u2 ) / ( 2 * a )
s = ( 0 - 2500 ) / ( 2 * -10 )
s = 125 metres
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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.
Where is the vertex coordinate of the parabola y equals 24 -6x -3x squared when plotted on the Cartesian plane?
The vertex coordinate point of the vertex of the parabola y = 24-6x-3x^2 when plotted on the Cartesian plane is at (-1, 27) which can also be found by completing the square.
To find the value of a in a parabola opening up or down, subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the of the vertex?
right
What is the average of the two roots of quadratic equation?
In a quadratic y = ax² + bx + c, the roots are where y = 0, and the parabola crosses the x-axis. The average of these two roots is the x coordinate of the vertex of the parabola.
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.
When a parabola opens upward the y coordinate of the vertex is a what?
Opening up, the vertex is a minimum.
What is the y-coordinate of the vertex of a parabola with the following equation?
The y coordinate is given below:
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.
To find the value of a in a parabola opening up or down subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the of the vertex?
To find the value of a in a parabola opening up or down subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the right of the vertex.
Where is the vertex coordinate of the parabola y equals x square -3x -10 located on the Cartesian plane?
7
Where is the vertex coordinate of the parabola y equals 24 -6x -3x squared when plotted on the Cartesian plane?
The vertex coordinate point of the vertex of the parabola y = 24-6x-3x^2 when plotted on the Cartesian plane is at (-1, 27) which can also be found by completing the square.
To find the value of a in a parabola opening up or down, subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the of the vertex?
right
What is the y-coordinate of the vertex of the parabola that is given by the equation below?
We will be able to identify the answer if we have the equation. We can only check on the coordinates from the given vertex.
What is the average of the two roots of quadratic equation?
In a quadratic y = ax² + bx + c, the roots are where y = 0, and the parabola crosses the x-axis. The average of these two roots is the x coordinate of the vertex of the parabola.
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.
What is the y coordinate of the vertex of the parabola that is given y equals negative 2 point 5 parenthese x minus 4 parenthese end squared minus 5?
y = -5 By using calculus, the derivative of y = -2.5(x-4)2 - 5 is y' = -5(x-4). Solving the equation -5(x-4) = 0 gives x = 4 (since the slope of the parabola at the vertex is zero). Plug this back into the equation: y = -2.5(4 - 4) -5 = -5, so the y-coordinate is -5. The equation of the parabola is given in the vertex form y = a(x - h)2 + k, where (h, k) is the vertex. So the vertex is (4, -5).
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).
