point of symmetry?
2(x-7)(X+3)
and vertex?
This is a parabola. The axis of symmetry is midway between the roots, or the line
x = 2
There is no point of symmetry, just an axis.
The vertex is at (2,-50)
but the anwer is (-3,0)(7,0)
how can i get that?
To find the point of symmetry and vertex of the equation 2(x-7)(x+3), which appears to be a quadratic equation, you can start by expanding the equation and putting it into general form.
Step 1: Expand the equation
Multiplying the terms 2(x-7) (x+3) will give you:
2(x^2 - 4x - 21)
Step 2: Simplify the equation
Distribute the 2 to each term in parentheses:
2x^2 - 8x - 42
The equation is now in standard quadratic form, which is ax^2 + bx + c. In this case, a = 2, b = -8, and c = -42.
Step 3: Find the axis of symmetry
The axis of symmetry is a vertical line that divides the parabola into two equal halves. In a quadratic equation of the form ax^2 + bx + c, the x-coordinate of the point of symmetry is given by the formula: x = -b / 2a.
In this case, -b = -(-8) = 8 and 2a = 2(2) = 4.
x = 8 / 4 = 2
So, the x-coordinate of the point of symmetry is 2.
Step 4: Find the y-coordinate of the vertex
To find the y-coordinate of the vertex, substitute the x-coordinate of the point of symmetry into the original equation.
2(2)^2 - 8(2) - 42 =
2(4) - 16 - 42 =
8 - 16 - 42 =
-50
So, the y-coordinate of the vertex is -50.
Therefore, the point of symmetry is (2, -50) and the vertex is located at (2, -50).