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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}))).
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What is the standard form of the equation of the parabola with vertex 00 and directrix y4?
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
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 write y equals x minus 4 x plus 2 in vertex form and find the vertex?
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.
The vertex form of the equation of a parabola is . What is the standard form of the equation?
To convert the vertex form of a parabola, which is typically expressed as (y = a(x-h)^2 + k), into standard form (y = ax^2 + bx + c), you need to expand the equation. Start by squaring the binomial ((x-h)), which gives (x^2 - 2hx + h^2). Then, distribute the coefficient (a) and combine like terms to achieve the standard form. The resulting equation will be (y = ax^2 - 2ahx + (ah^2 + k)).
The vertex form of the equation of a parabola is y 4(x - 2)2 - 1. What is the standard form of the equation?
To convert the vertex form ( y = 4(x - 2)^2 - 1 ) to standard form, expand the equation. Start by expanding ( (x - 2)^2 ) to get ( x^2 - 4x + 4 ). Then, substitute this back into the equation: ( y = 4(x^2 - 4x + 4) - 1 ), which simplifies to ( y = 4x^2 - 16x + 16 - 1 ). Therefore, the standard form is ( y = 4x^2 - 16x + 15 ).
What is the difference between standard form and vertex form?
The difference between standard form and vertex form is the standard form gives the coefficients(a,b,c) of the different powers of x. The vertex form gives the vertex 9hk) of the parabola as part of the equation.
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.
What different information do you get from vertex form and quadratic equation in standard form?
The graph of a quadratic function is always a parabola. If you put the equation (or function) into vertex form, you can read off the coordinates of the vertex, and you know the shape and orientation (up/down) of the parabola.
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)
What is the equation for vertex form?
The vertex form for a quadratic equation is y=a(x-h)^2+k.
What is the standard form of the equation of the parabola with vertex 00 and directrix y4?
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
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)).
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
The vertex of the parabola below is at the point (-4-2) which equation below could be one for parabola?
-2
How do you write y equals x minus 4 x plus 2 in vertex form and find the vertex?
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.
The vertex form of the equation of a parabola is y 4(x - 2)2 - 1. What is the standard form of the equation?
To convert the vertex form ( y = 4(x - 2)^2 - 1 ) to standard form, expand the equation. Start by expanding ( (x - 2)^2 ) to get ( x^2 - 4x + 4 ). Then, substitute this back into the equation: ( y = 4(x^2 - 4x + 4) - 1 ), which simplifies to ( y = 4x^2 - 16x + 16 - 1 ). Therefore, the standard form is ( y = 4x^2 - 16x + 15 ).
How do you find the vertex of an equation in vertex 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)
