Hello!
First, let's rewrite the expression here:
[tex]f\mleft(x\mright)=-2x^2+12x-15[/tex]
We will have to find the coefficients a, b and c:
• a ,= -2;
,
• b ,= 12;
,
• c ,= -15;
As we can see, coefficient a is negative. It means that the parabola will face downwards and have a maximum value.
We can obtain this point by using the two formulas below to calculate the coordinates (x, y) of this point, look:
[tex]\begin{gathered} X_V=\text{ }-\frac{b}{2\cdot a} \\ \\ Y_V=-\frac{b^2-4\cdot a\cdot c}{4\cdot a} \end{gathered}[/tex]
As we know the coefficients, let's replace the values in the formulas:
Xv:
[tex]\begin{gathered} X_V=-\frac{b}{2\cdot a} \\ \\ X_V=-\frac{12}{2\cdot(-2)}=-\frac{12}{-4}=-(-3)=+3 \\ \\ X_V=3 \end{gathered}[/tex]
Now let's find Yv:
[tex]\begin{gathered} Y_V=-\frac{b^2-4\cdot a\cdot c}{4\cdot a} \\ \\ Y_V=-\frac{12^2-4\cdot(-2)\cdot(-15)}{4\cdot(-2)}=-\frac{144-120}{-8}=-\frac{24}{-8}=-(-3)=+3 \\ \\ Y_V=3 \end{gathered}[/tex]
Doing this, we obtained the coordinate of the maximum point (x, y) = (3, 3).
As it doesn't have any restrictions, the domain will be: [-∞, +∞].
[tex]-\infty\: We
have one restriction in the range, do you remember which?
The maximum value will be when y = 3, so the range is [-∞, 3].
Look at the graph of this function below:
