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how do u find the x-intercepts of a quadratic function algebraically?
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let y=0
then solve for x
k thanks!
yw
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Or for \[y = ax^{2} + bx + c\] intercepts will occur at: \[\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\]
algebraically? cant be down
there has to be a full moon, dancing, and a sacrificed chicken .... only then will the gods betow upon you the answers to the quadratics
...and maybe even a spellcheck lol
The long way would be: \[0=ax^{2}+bx+c\] \[0=4(ax)^{2}+4abx+4ac\] \[-4ac=4(ax)^{2}+4abx\] \[b^{2}-4ac=4(ax)^{2}+4abx+b^{2}\] \[b^{2}-4ac=(2ax+b)^{2}\] \[\pm\sqrt{b^{2}-4ac}=2ax+b\] \[-b\pm\sqrt{b^{2}-4ac}=2ax\] \[\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=x\]
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