(0, 0), of course. No linear term.
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.
The vertex is the origin, (0,0).
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.
x2 * -3x = -3x3
To determine if a vertex is a minimum in a quadratic function, you can analyze the coefficient of the quadratic term (the leading coefficient). If the coefficient is positive, the parabola opens upwards, indicating that the vertex is a minimum point. Additionally, you can use the second derivative test; if the second derivative at the vertex is positive, the vertex is confirmed as a minimum.
(0, 0) of course! No linear term! Review you vertex manipulation again.
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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.
The vertex is the origin, (0,0).
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.
x2 * -3x = -3x3
To determine if a vertex is a minimum in a quadratic function, you can analyze the coefficient of the quadratic term (the leading coefficient). If the coefficient is positive, the parabola opens upwards, indicating that the vertex is a minimum point. Additionally, you can use the second derivative test; if the second derivative at the vertex is positive, the vertex is confirmed as a minimum.
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
To find the vertex of the quadratic equation ( y = 3x^2 - 12x - 5 ), we can use the vertex formula ( x = -\frac{b}{2a} ), where ( a = 3 ) and ( b = -12 ). Plugging in the values, we get ( x = -\frac{-12}{2 \cdot 3} = 2 ). To find the corresponding ( y )-coordinate, substitute ( x = 2 ) back into the equation: ( y = 3(2)^2 - 12(2) - 5 = -29 ). Thus, the coordinates of the vertex are ( (2, -29) ).
y = 3x2+2x-1 Line of symmetry: x = -1/3 Vertex coordinate: (-1/3, -4/3)
x(3x+6) or 3x(x+2). Id recomend the second one.
let the vertex angle be x degrees, then the base angle is x + 9 degrees. Since in a triangle the sum of the angle is 180 degrees, and the base angles in an isosceles triangle are congruent, we have: x + 2(x + 9) = 180 x + 2x + 18 = 180 3x + 18 = 180 subtract 18 to both sides 3x = 162 divide by 3 to both sides x = 54 Thus the vertex angle is 54 degrees.