Given the function f(x)= 4x^2+32x+50 find the vertex of the graph of f(x), find the x intercepts if any, find the maximum value obtained by f(x), find minimum

Question

 Please help I am at a loss with this problem

Answer 1

First, we can put the given function in vertex form.  This form will make everything easier to find, such as intercepts and minimums and maximums.

 

y = a(x - h)² + k

 

where:

a is the coefficient of the x² term

vertex has the coordinates (h, k).

 

 

y = 4(x² + 8x + 16) - 14

y = 4(x + 4)(x + 4) - 14

y = 4(x + 4)² - 14

 

h = -4

k = -14

 

The vertex is (-4, -14).  This coordinate is also the maximum or minimum value. Since the first term of the function is positive, and the function is a parabola, we only have a minimum. 

 

Therefore, (4, -14) is the minimum.

 

 

To find the x-intercept, plug in y=0 into the vertex form of the function and solve for x.

To find the y-intercept, plug in x=0 into the vertex form of the function and solve for y.

 

Answer 2

f(x) = 4x² + 32x + 50

 

Vertex is at (h,K) where h = -b/2a

-b/2a = -32/2(4) = -32/8 = -4

The x coordinate of the vertex is -4

k = f(-4) = 4(-4)² + 32(-4) + 50 = 64 -128 +50 = -14

 

The vertex is at (-4,-14)

 

As noted, since the coefficient of the squared term is positive the parabola opens upward

Thus the minimum value of f(x) is the y value of the vertex

Minimum value is -14

 

Since the parabola opens upward it has no maximum value.... infinity

 

As for the x-intercepts, set the equation equal to zero and solve for x

 

4x² + 32x + 50 = 0

 

We can factor out a 2 and work with smaller numbers

2(2x²+16x+25)=0

2x²+16x+25 = 0

 

a = 2, b=16, b=25   x=-b/2a ±√(b²-4ac)/2a

 

x = -16/2(2) ±[√(16²-4(2)(25)]/2(2)

 = -4 ± [√(256-200)]/4

 = -4 ± (√56)/4

 = -4 ± 2√14/4

 = -4 ± √14/2

 

The x intercepts are at (-4+√14/2, 0)   and  (-4-√14/2, 0)

 

Lets check one of the answers:

 

f(-4+√14/2) = 0?

 

4x²+32x+50 = 4(-4+√14/2)² + 32(-4 +√14/2) + 50

 

= 4(16 -4√14 + 14/4) - 128 + 16√14 + 50

 

= 64 -16√14 + 14 - 128 + 16√14 + 50

 

= 128 - 128 = 0

 

This answer checks.

 

Answer 3

4x² + 32x + 50    The x-value of the vertex can be found from the formula "-b/2a" , which in this case is -32/(2*4) = -4 The y-value is obtained by plugging in -4 for x:     4(-4)² + 32(-4) + 50 = -14. So vertex is (-4,-14)

The minimum value is also -14, because this parabola opens up, since the coefficient of x² is positive.

 

Since it opens upwards, the maximum value of the function is "infinity".

 

The x-intercepts occur when the function crosses the x-axis. Here, the value of f(x) [or y, if you like] = 0

 

So 4x² + 32x + 50 =0 If we divide by 2, we get: 2x² + 16x + 25=0  

 

This won't factor, so we solve using the quadratic formula:

 

x = {-16 ± √[(16²) -4(2)(25)] }/2(2)  

From here, we get x = { -16 ± √56 }/4 = -4 ± √14