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The quadratic equation, in its standard form is: ax2 + bx + c = 0 where a, b and c are constants and a is not zero.
it is 1
No, it is not.
In the beginning, no you need not cull the special coefficient
Put the quadratic equation into standard form; identify the coefficients (a, b, c), replace them in the equation, do the calculations.
The quadratic equation, in its standard form is: ax2 + bx + c = 0 where a, b and c are constants and a is not zero.
ax2 +bx + c = 0
The slope of your quadratic equation in general form or standard form.
it is 1
The question i have to convert to standard form is -1/2(x-6)2
ax^2+bx+c=0 is the standard form of a quadratic function.
ax2 + bx + c = 0 where a, b and c are constants and a is not 0.
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.
It is still called a quadratic equation!
Normally a quadratic equation will graph out into a parabola. The standard form is f(x)=a(x-h)2+k
Standard notation for a quadratic function: y= ax2 + bx + c which forms a parabola, a is positive , minimum value (parabola opens upwards on an x-y graph) a is negative, maximum value (parabola opens downward) See related link.
No, it is not.