f(x)=−4x2−2x+1 For the quadratic function: a) Find the equation of the axis of symmetry: b) Find the vertex: c) Find the intercepts and express them as points.

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f(x)=−4x2−2x+1  For the quadratic function:  a) Find the equation of the axis of symmetry:  b) Find the vertex:  c) Find the intercepts and express them as points.

SchulzD · Accepted Answer

Step 1  The quadratic equation  f(x)=4x2−2x+1  We have to solve for part (a), (b) and (c).  Step 2  a) We have  f(x)=4x2−2x+1  We can write  y=4x2−2x+1  ⇒y=4x2−2x+14+1−14  (added and subtracted 14)  ⇒y=(2x−12)2+34  ⇒(y−34)=(2x−12)2  Step 3  Now axis of symmetry can found as by making (2x−12)2=0  ⇒(2x−12)2=0  ⇒(2x−12)=0  ⇒2x=12  ⇒x=14  So, the equation axis of symmetry is x=14  Step 4  b) Vertex can be found as by making (2x−12)2=0 and (y−34)=0  ⇒(2x−12)=0 and (y−34)=0  ⇒2x=12 and y−34=0  ⇒x=14 and y=34  So, the vertex of parabola is (14, 34)  Step 5  c) For y intercept; x=0  ⇒y=4×02−2×0+1  ⇒y=0−0+1  ⇒y=1  So, y intercept is (0, 1)  For x intercept; y=0  ⇒0=(2x−12)2+34  ⇒(2x−12)2=−34  But square of any number cannot be negative,  So, x intercept will not be there.

f(x)=-4x^{2}-2x+1 For the quadratic function: a) Find the equation

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