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what is the solution set for x^2-8x+4>-3?
x<1 or x>7
7
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x^2 - 8x + 4 > - 3 Add 3 to both sides: x^2 - 8x + 4 + 3 > 0 x^2 - 8x + 7 > 0
The left side factors to (x - 7)(x - 1) > 0 Find the value of the vertex \[\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)\] Then graph x^2 - 8x + 7
From there, you can easily observe the parts where the graph is greater than zero
Thank you ( : does this work for all problems like this?
what is the variable f? @Hero
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For this set of problems, yes.
We can refer to the quadratic as f(x) = x^2 - 8x - 7
Once you find the x coordinate of the vertex, you substitute it in for x, then evaluate
Have you found the x coordinate for the vertex yet?
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