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}))).
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
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)).
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.
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 ).
To convert a quadratic equation from vertex form, (y = a(x - h)^2 + k), to standard form, (y = ax^2 + bx + c), you need to expand the equation. Start by squaring the binomial: ( (x - h)^2 = x^2 - 2hx + h^2 ). Then, multiply by (a) and add (k) to obtain (y = ax^2 - 2ahx + (ah^2 + k)), where (b = -2ah) and (c = ah^2 + k). This results in the standard form of the quadratic equation.
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.
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.
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.
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)
The vertex form for a quadratic equation is y=a(x-h)^2+k.
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
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)).
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
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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.
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 ).
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)