3+4x-x^2=h(x)

Question

Write the quadratic function in standard form identify vertex axis O's symmetry and x intercepts

Answer 1

Hi Gj,

 

The standard form of an quadratic equation is f(x) = a(x - h)2 + k

 

We have 3 + 4x - x² = h(x)

 

Rearranging terms

h(x) = -(x²-4x-3)

 

Let's ignore the - for a moment and we have

 

x²-4x-3

 

If we set this equal to 0, we have

x²-4x-3 = 0

or

x²-4x = 3

 

Now complete the squares

x²-4x+4 = 3+4 = 7

(x-2)² = 7

or

(x-2)²- 7 = 0

 

Here 2 is the h and -7 is the k

 

This is almost the standard form but we have to remember the "-" we put aside before.

 

Putting it back in,

-[(x-2)²- 7] = h(x)

and distributing the "-"

 

h(x) = -(x-2)²+7

 

The vertex is the point (h,k), which is (2, -7)

The axis of symmetry is x=2

 

Finally, the x intercepts are the point where y=0

Going back to the equation as originally stated

h(x) = -x2 + 4x + 3

Set this equal to 0 since we want the x's when h(x) = 0

0 = -x² + 4x + 3

Let's divide by -1 so that the leading coefficient is +1, not -1

0 = x²-4x-3

 

We can't solve this by factoring but we can use the quadratic formula

-b±√b²-4ac

-------------

      2a

 

+1±√[(-1)2-4(1)(-3)]

-------------------------

          2(1)

 

1 ±√(1+12)

-------------

        2

 

1±√13

-------

   2

 

Some teachers want answers left as radicals while others want exact answers.

 

1±3.60555

------------

      2

 

-1.3, 2.3 <---x-intercepts

 

Answer 2

y - 3 = -(x²- 4x   )

y - 3 - 4 = -(x² - 4x + 4)

y - 7 = -(x - 2)²

 

Vertex: (2,7)

 

Axis of symmetry: x = 2

 

x-intercepts: 2 ± √7 using quadratic formula.