Question

In the xy-coordinate plane, the graph of the equation y=2x^2-12x-32 has zeros at x=d and x=e, where d>e. The graph has a minimum at (f,-50) what are the vaules of d,e,and f?

Answer 1

to find the zeroes  solve fro x:-

2(x^2 - 6x - 16) = 0

2(x - 8)(x + 2) = 0

x =  {8, -2
 so d = 8 and e = -2

the x coordinate of the minimum value = - b / 2a =  12/ 2*2  =  3

f = 3

Answer 2

Answer:  The values are d = 8, e = - 2  and  f = 3.

Step-by-step explanation:  Given that in the xy-plane, the graph of the equation $y=2x^2-12x-32$ has zeros at x=d and x=e, where d > e. The graph has a minimum at (f, -50).

We are to find the values of d, e and f.

The zeroes of the given function are

$2x^2-12x-32=0\\\\\Rightarrow x^2-6x-16=0\\\\\Rightarrow x^2-8x+2x-16=0\\\\\Rightarrow x(x-8)+2(x-8)=0\\\\\Rightarrow (x-8)(x+2)=0\\\\\Rightarrow x-8=0,~~~~~x+2=0\\\\\Rightarrow x=8,~~~~~~~~\Rightarrow x=-2.$

So, d = 8 and e = -2.

We know that the minimum value of a function $y=ax^2+bx+c$ occurs at $x=-\dfrac{b}{2a}.$

Therefore, the given function will have a minimum values at

$x=-\dfrac{b}{2a}=-\dfrac{-12}{2\times 2}=\dfrac{12}{4}=3.$

Therefore, the function has minimum at (3, -50) and so the value of f is 3.

Thus, d = 8, e = - 2  and  f = 3.