First the formula is g(x)=ax2+bx+c First find where the parabola cuts the x axis Then find the equation of the axis of symmetry Then
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y = 2x + 2 + 4x+ 2 = 6x + 4 This is NOT a symmetric function and so there is no axis of symmetry.
To find the axis of symmetry for the quadratic equation ( y = -x^2 + 2x - 4 ), you can use the formula ( x = -\frac{b}{2a} ), where ( a ) and ( b ) are the coefficients from the equation in standard form ( y = ax^2 + bx + c ). Here, ( a = -1 ) and ( b = 2 ). Plugging in the values, the axis of symmetry is ( x = -\frac{2}{2 \times -1} = 1 ). Thus, the axis of symmetry is ( x = 1 ).
To determine if a graph is symmetric with respect to the x-axis, check if replacing (y) with (-y) in the equation yields an equivalent equation. For y-axis symmetry, replace (x) with (-x) and see if the equation remains unchanged. For origin symmetry, replace both (x) with (-x) and (y) with (-y) and verify if the equation is still the same. If the equation holds true for any of these conditions, the graph exhibits the corresponding symmetry.
Your equation must be in y=ax^2+bx+c form Then the equation is x= -b/2a That is how you find the axis of symmetry
If a quadratic function has the points (-4,0) and (14,0), what is equation of the axis of symmetry?
First the formula is g(x)=ax2+bx+c First find where the parabola cuts the x axis Then find the equation of the axis of symmetry Then
There is no equation (nor inequality) in the question so there can be no graph - with or without an axis of symmetry.
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X=-b/2a
y = 2x + 2 + 4x+ 2 = 6x + 4 This is NOT a symmetric function and so there is no axis of symmetry.
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
Complete the square, then find the value of x that would make the bracket zero ax^2 + bx + c = 0 line of symmetry is x = (-b/2a)
12 x 5 = 60
Line of symmetry: x = 3
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