# What the minimum of { y=2x^2-7x-2 }? How do you find it?

## Question:

What the minimum of {eq}y=2x^2-7x-2 {/eq}? How do you find it?

## Quadratic Function and its Minimum Value:

{eq}\\ {/eq}

The standard form of the quadratic function {eq}\Biggr[ f(x) = ax^{2} + bx + c \Biggr] {/eq} will either have the maximum value or minimum value, it all depends on the value of the coefficients of {eq}x^{2} {/eq} or on the value of {eq}a {/eq}. If the value of {eq}a {/eq} is positive then the function will have only a minimum value. If the value of {eq}a {/eq} is negative then the function will have only a maximum value. Here we will use the derivatives (first and second) in order to get the minimum value of the function. First of all, we will determine the critical points then we will check at which point, the function will have a point of minima using the second derivative test.

$$\text {Critical Points} \; \; \; \Longrightarrow \; \; \biggr( \dfrac {dy}{dx} \biggr) = 0 $$

$$\text {Point of Minima} \; \; \; \Longrightarrow \; \; \; \biggr( \dfrac {d^{2}y}{dx^{2}} \biggr) > 0 $$

## Answer and Explanation:

{eq}\\ {/eq}

The function for which we have to determine the minimum value is given below:

{eq}y = 2x^{2} - 7x - 2 {/eq}

Now determine the critical points of the function using first derivative:

{eq}\biggr( \dfrac {dy}{dx} \biggr) = \dfrac {d(2x^{2} - 7x - 2)}{dx} {/eq}

{eq}\biggr( \dfrac {dy}{dx} \biggr) = \dfrac {d(2x^{2})}{dx} - 7 \dfrac {d(x)}{dx} - \dfrac {d(2)}{dx} {/eq}

{eq}\biggr( \dfrac {dy}{dx} \biggr) = 4x - 7 {/eq}

{eq}\biggr( \dfrac {dy}{dx} \biggr) = 0 {/eq}

{eq}4x - 7 = 0 \; \; \; \Longrightarrow \; \; x = \biggr( \dfrac {7}{4} \biggr) {/eq}

Now we will check that the point {eq}x = \biggr( \dfrac {7}{4} \biggr) {/eq} will either lead to a point of maxima or minima:

{eq}\biggr( \dfrac {d^{2}y}{dx^{2}} \biggr) = 4 > 0 {/eq}

Hence the point {eq}x = \biggr( \dfrac {7}{4} \biggr) {/eq} will lead to the point of minima:

{eq}f \biggr( \dfrac {7}{4} \biggr) = 2 \biggr( \dfrac {7}{4} \biggr)^{2} - 7 \biggr( \dfrac {7}{4} \biggr) - 2 {/eq}

{eq}f \biggr( \dfrac {7}{4} \biggr) = \dfrac {49}{8} - \dfrac {49}{4} - 2 = - \biggr( \dfrac {65}{8} \biggr) {/eq}

Hence the minimum value of the function {eq}y = f(x) {/eq} is {eq}- \biggr( \dfrac {65}{8} \biggr) {/eq}.

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from Math 104: Calculus

Chapter 9 / Lesson 4