Determine the vertex and axis of symmetry of the following quadratics. a) f(x)=2(x−3)2−1 b) f(x)=x2+4x+3

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Determine the vertex and axis of symmetry of the following quadratics.  a) f(x)=2(x−3)2−1  b) f(x)=x2+4x+3

Esta Hurtado · Accepted Answer

Step 1  For a quadratic equation y=ax2+bx+c  Axis of Symmetry is x=−b2a  x-coordinate of vertex is −b2a  a) f(x)=2(x−3)2−1  =2(x2−6x+9)−1  f(x)=2x2−12x+17  Axis of symmetry ⇒x=−b2a  ⇒x=−(−12)2(2)=3  ⇒x=3  Step 2  x-coordinate of vertex =−b2a=3  y-coordinate of vertex =f(−b2a)=f(3)  =2(3−3)2−1=−1  ∴ Vertex =(3, −1)  b) f(x)=x2+4x+3  Axis of symmetry  ⇒x=−b2a=−42(1)=−2  ⇒x=−2  x-coordinate of vertex =−b2a=−2  Step 3  y-coordinate of vertex  =f(−b2a)  =f(−2)  =(−2)2+4(−2)+3  =−1  ∴ Vertex =(−2, −1)

Donald Cheek · Accepted Answer

Step 1  f(x)=2(x−3)2−1  Multiply 2 and 2 to get 4.  4(x−3)−1  Use the distributive property to multiply 4 by x−3.  4x−12−1  Subtract 1 from -12 to get -13  4x−13  Step 2  f(x)=x2+4x+3  Factor the expression by grouping. First, the expression needs to be rewritten as x2+ax+bx+3. To find a and b, set up a system to be solved  a+b=4  ab=1×3=3  Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.  a=1  b=3  Rewrite x2+4x+3 as (x2+x)+(3x+3)  (x2+x)+(3x+3)  Factor out x in the first and 3 in the second group.  x(x+1)+3(x+1)  Factor out common term x+1 by using distributive property.  (x+1)(x+3)

Determine the vertex and axis of symmetry of the following

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