Quadratic Functions
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
Solve by factorising:
(a)\(x^2+5x+6=0\)
(b)\(x^2-7x+12=0\)
(c)\(3x^2+5x-2=0\)
(a)
Factorise
\((x+2)(x+3)=0\)
Answer
\(x=-2\) or \(x=-3\)
(b)
Factorise
\((x-3)(x-4)=0\)
(c)
Factorise
\((3x-1)(x+2)=0\)
Answer
\(x= frac{1}{3}\) or \(x=-2\)
Question 2
Solve using the quadratic formula:
(a)\(3x^2 + 7x - 5 = 0\). Give your answer to 3 s.f.
(b)\(2x^2 - 4x - 1 = 0\). Leave in surd form.
(a)
Formula
\(x=\dfrac{-7\pm\sqrt{49+60}}{6}=\dfrac{-7\pm\sqrt{109}}{6}\)
Answer
\(x=0.573\) or \(x=-2.91\)
(b)
Formula
\(x=\dfrac{4\pm\sqrt{16+8}}{4}=\dfrac{4\pm\sqrt{24}}{4}\)
Answer
\(x=\dfrac{2\pm\sqrt{6}}{2}\)
Question 3
Rewrite the following in the form \(y=r(x-p)^2+q\) and state the vertex:
(a)\(y=x^2+4x+1\)
(b)\(y=x^2-6x+11\)
(c)\(y=2x^2+8x-3\)
(a)
Answer
\(y=(x+2)^2-3\); vertex \((-2,-3)\)
(b)
Answer
\(y=(x-3)^2+2\); vertex \((3,2)\)
(c)
Factor out 2
\(2(x^2+4x)-3=2(x+2)^2-8-3\)
Answer
\(y=2(x+2)^2-11\); vertex \((-2,-11)\)
Question 4
Sketch the following quadratics, showing coordinates of the vertex and \(y\)-axis intercept:
(a)\(y=(x+1)^2-4\)
(b)\(y=-2(x-3)^2+5\)
(c)\(y=3(x+2)(x-1)\)
(a)
Answer
Vertex \((-1,-4)\); \(y\)-int: \((0,-3)\); opens up; roots at \(x=1,-3\)
(b)
Answer
Vertex \((3,5)\); \(y\)-int: \((0,-13)\); opens down
(c)
Answer
Roots at \(x=-2,1\); vertex \(x=- frac{1}{2}\); \(y\)-int: \((0,-6)\)
Question 5
Find the discriminant and state the nature of the roots:
(a)\(x^2-6x+9=0\)
(b)\(2x^2+3x+5=0\)
(c)\(x^2-5x+4=0\)
(a)
Answer
One repeated real root: \(x=3\)
(c)
Answer
Two distinct real roots
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