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SAT II Math II : Maximum and Minimum

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

Example Question #1 : Maximum And Minimum

What is the vertex of y=-3x^2-5x+3 ? Is it a max or min?

Possible Answers:

$\frac{5}{6}, \textup{Min}$

$-\frac{5}{6}, \textup{Max}$

$-\frac{5}{6}, \textup{Min}$

$-\frac{5}{3}, \textup{Max}$

$-\frac{5}{3}, \textup{Min}$

Correct answer:

$-\frac{5}{6}, \textup{Max}$

Explanation:

The polynomial is in standard form of a parabola.

y=ax^2+bx+c

To determine the vertex, first write the formula.

$x=-\frac{b}{2a}$

Substitute the coefficients.

$x=-\frac{-5}{2(-3)} = -\frac{5}{6}$

Since the is negative is negative, the parabola opens down, and we will have a maximum.

The answer is: $-\frac{5}{6}, \textup{Max}$

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Example Question #1 : Maximum And Minimum

Given the parabola equation y=4x-3x^2 , what is the max or minimum, and where?

Possible Answers:

$\textup{Min}, (-\frac{2}{3},-4)$

$\textup{Max}, (\frac{2}{3},4)$

$\textup{Max}, (-\frac{2}{3},-4)$

$\textup{Max}, (\frac{2}{3}, \frac{4}{3})$

$\textup{Min}, (\frac{2}{3}, \frac{4}{3})$

Correct answer:

$\textup{Max}, (\frac{2}{3}, \frac{4}{3})$

Explanation:

The parabola is in the form: y=ax^2+bx+c

y=-3x^2+4x

The vertex formula will determine the x-value of the max or min. Since the value of is negative, the parabola will open downward, and there will be a maximum.

Write the vertex formula and substitute the correct coefficients.

$x=-\frac{b}{2a} = -\frac{4}{2(-3)} = \frac{2}{3}$

Substitute this value back in the parabolic equation to determine the y-value.

$y=-3( \frac{2}{3})^2+4( \frac{2}{3}) = \frac{4}{3}$

The answer is: $\textup{Max}, (\frac{2}{3}, \frac{4}{3})$

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All SAT II Math II Resources

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SAT II Math II : Maximum and Minimum

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 : Maximum And Minimum

Example Question #1 : Maximum And Minimum

All SAT II Math II Resources

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