Question

Given y = -8x^2 - 64x + 72, what is the highest possible value y can have?

Answer 1

- The equation given is a quadratic equation and the graph of a quadratic equation has either a minimum or a maximum never both.

- The equation given is

`y=-8x^2-64x+72`

- This equation has a maximum value because:

`\begin{gathered} Given\text{ a quadratic equation,} \\ ax^2+bx+c \\ \\ if\text{ }a<0,\text{ The equation has a maximum value} \\ if\text{ }a>0,\text{ The equation has a minimum value} \\ \\ \text{ Since }a=-8,\text{ we can be certain that the equation has a maximum value} \end{gathered}`

- The maximum value happens to coincide with the vertex of the equation. The x-value for this vertex is gotten using the formula below:

`\begin{gathered} x=-(b)/(2a) \\ \\ \text{ For the equation,} \\ y=ax^2+bx+c \end{gathered}`

- Thus, we can find the x-value of the vertex using the above formula. Once we have the x-value of the vertex, we can simply substitute the x-value into the equation to get the corresponding y-value which would represent the highest possible y-value.

- Thus, we have:

`\begin{gathered} a=-8,b=-64,c=72 \\ x=-(b)/(2a)=-(-64)/(2(-8))=-4 \\ \\ \\ \text{ Put the value of x into the equation, we have:} \\ y=-8x^2-64x+72 \\ y=-8(-4)^2-64(-4)+72 \\ y=-128+256+72 \\ y=200 \end{gathered}`

Final Answer

The highest possible y-value is y = 200