Quadratic Functions
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jonah2.0
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Beer soaked ramblings follow.
Hi guys I have this homework problem maybe you could help me out
Determine an equation for the quadratic function with zeros of -2 + sqrt5 and -2 - sqrt5, containing the point (-4,5). Write the equation in factored, and standard form.
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Harry_the_cat
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If the zeroes of a quadratic function are p and q, then the factors are (x-p) and (x-q).Hi guys I have this homework problem maybe you could help me out
Determine an equation for the quadratic function with zeros of -2 + sqrt5 and -2 - sqrt5, containing the point (-4,5). Write the equation in factored, and standard form.
You are given p and q, so you should be able to form the function as y = a(x-p)(x-q).
Then sub in the point (-4,5) to find a.
Give it a go and see what you can do. Come back if you need further help.
HallsofIvy
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You can write a quadratic function as \(\displaystyle f(x)= a(x- x_0)(x- x_1)\) where a is some number and \(\displaystyle x_0\) and \(\displaystyle x_1\) are values of x such that f(x)= 0.
So we can write \(\displaystyle f(x)= a(x- (-2- \sqrt{5}))(x- (-2+ \sqrt{5}))\). The multiplication is simplified if we use the fact that \(\displaystyle (p- q)(p+ q)= p^2- q^2\) with \(\displaystyle p= x+ 2\) and \(\displaystyle q= \sqrt{5}\(\displaystyle .
So \(\displaystyle f(x)= a(x- (-2- \sqrt{5}))(x- (-2+ \sqrt{5}))= a((x+ 2)^2- (\sqrt{5})^2)= a(x^2+ 4x+ 4- 5)= a(x^2+ 4x- 1)\). All that is left is to determine a. To do that, use the fact that \(\displaystyle f(-4)= a((-4)^2+ 4(-4)- 1)= a(16- 16- 1)= -a= 5.\)\)\)
So we can write \(\displaystyle f(x)= a(x- (-2- \sqrt{5}))(x- (-2+ \sqrt{5}))\). The multiplication is simplified if we use the fact that \(\displaystyle (p- q)(p+ q)= p^2- q^2\) with \(\displaystyle p= x+ 2\) and \(\displaystyle q= \sqrt{5}\(\displaystyle .
So \(\displaystyle f(x)= a(x- (-2- \sqrt{5}))(x- (-2+ \sqrt{5}))= a((x+ 2)^2- (\sqrt{5})^2)= a(x^2+ 4x+ 4- 5)= a(x^2+ 4x- 1)\). All that is left is to determine a. To do that, use the fact that \(\displaystyle f(-4)= a((-4)^2+ 4(-4)- 1)= a(16- 16- 1)= -a= 5.\)\)\)