1) Substitute 0 for x in the function:
f ( 0 ) = 0 2 − 0 − 12 = − 12 f(0)={{0}^{2}}-0-12=-12 f ( 0 ) = 0 2 − 0 − 12 = − 12 so
f ( 0 ) = − 12 f\left( 0 \right)=-12 f ( 0 ) = − 12 2) Solve
f ( x ) = 0 f\left( x \right)=0 f ( x ) = 0 We have equation
x 2 − x − 12 = 0 {{x}^{2}}-x-12=0 x 2 − x − 12 = 0 In order to solve this equation we use the quadratic formula. If
a x 2 + b x + c = 0 a{{x}^{2}}+bx+c=0 a x 2 + b x + c = 0 then
x = − b ± b 2 − 4 a c 2 a x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} x = 2 a − b ± b 2 − 4 a c In our case
a = 1 , b = − 1 , c = − 12 a=1, b=-1, c=-12 a = 1 , b = − 1 , c = − 12 Substitute these numbers in the formula
x = − ( − 1 ) ± ( − 1 ) 2 − 4 ⋅ 1 ⋅ ( − 12 ) 2 ⋅ 1 x=\frac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\cdot 1\cdot \left( -12 \right)}}{2\cdot 1} x = 2 ⋅ 1 − ( − 1 ) ± ( − 1 ) 2 − 4 ⋅ 1 ⋅ ( − 12 )
x = 1 ± 1 + 48 2 x=\frac{1\pm \sqrt{1+48}}{2} x = 2 1 ± 1 + 48
x = 1 ± 49 2 = 1 ± 7 2 x=\frac{1\pm \sqrt{49}}{2}=\frac{1\pm 7}{2} x = 2 1 ± 49 = 2 1 ± 7
x 1 = 1 + 7 2 = 4 , x 2 = 1 − 7 2 = − 3 {{x}_{1}}=\frac{1+7}{2}=4,\,\,\,\,{{x}_{2}}=\frac{1-7}{2}=-3 x 1 = 2 1 + 7 = 4 , x 2 = 2 1 − 7 = − 3 So our solution is
x 1 = 4 , x 2 = − 3 {{x}_{1}}=4,\,\,{{x}_{2}}=-3 x 1 = 4 , x 2 = − 3
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