Lg(5x+10)+2lg3=1+lg(4x+12)need them fast with steps
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if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.
so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.
so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.
or using a unit vector for those above, then
Answer:
In a quadratic equation of the shape:
y = a*x^2 + b*x + c
we hate that the discriminant is equal to:
D = b^2 - 4*a*c
This thing appears in the Bhaskara's formula for the roots of the quadratic equation:
You can see that the determinant is inside a square root, this means that if D is smaller than zero we will have imaginary roots (the graph never touches the x-axis)
If D = 0, the square root term dissapear, and this implies that both roots of the equation are the same, this means that the graph touches the x axis in only one point, wich coincides with the minimum/maximum of the graph)
If D > 0 we have two different roots, so the graph touches the x-axis in two different points.