Question Video: Finding the Maximum or Minimum Value of a Quadratic Function | Nagwa Question Video: Finding the Maximum or Minimum Value of a Quadratic Function | Nagwa

# Question Video: Finding the Maximum or Minimum Value of a Quadratic Function Mathematics

Find the maximum value of the function π¦ = β2π₯Β² + 12π₯ β 25.

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### Video Transcript

Find the maximum value of the function π¦ is equal to negative two π₯ squared plus 12π₯ minus 25.

In this question, weβre given a function π¦ which is a quadratic function. And weβre asked to find the maximum value of this function. We might be tempted to try applying the extreme value theorem to this question. However, the extreme value theorem is not applicable because weβre not looking for a maximum value over a closed interval. Instead, we need to find the maximum value of this function over the entire set of real numbers.

To do this, letβs start by considering a sketch of our function. Since our function is a quadratic with a negative leading coefficient, we know it will have a parabola shape facing downwards. And the maximum value of this function will be at the vertex or turning point of this function. And in fact, we can prove this result and find this value in two different ways. We can either complete the square or we can use calculus. Letβs start by using calculus. First, we recall the local extrema of a function must occur at their critical points. And we also recall the critical points of a function are where the derivative is equal to zero or where the derivative does not exist.

And in our case, because π¦ is a quadratic function, we know itβs continuous for all real values of π₯ and differentiable for all real values of π₯. So the only critical point of this function will be where its derivative is equal to zero, the maximum value of our function. So to find the maximum value of this function, we need to differentiate this with respect to π₯. We can do this term by term by using the power rule for differentiation. We multiply by the exponent of π₯ and reduce this exponent by one. Applying this to the first term, we get negative four π₯. And although we could apply this to evaluate the derivative of the next two terms, we notice these two terms are a linear function.

And we recall the derivative of a linear function is the coefficient of the variable, in this case 12. dπ¦ by dπ₯ is negative four π₯ plus 12. Now, weβre ready to find the only critical point of our function. It will be when the derivative is equal to zero. So we need to solve the equation negative four π₯ plus 12 is equal to zero. We subtract 12 from both sides of the equation and divide through by negative four to get that π₯ is equal to three. Finally, weβre asked to find the maximum value of this function. So we need to substitute π₯ is equal to three into our function. This gives us that π¦ is equal to negative two times three squared plus 12 times three minus 25.

And if we evaluate this expression, we see itβs equal to negative seven. The maximum value of the function π¦ is equal to negative two π₯ squared plus 12π₯ minus 25 is negative seven. However, this is not the only way we could have answered this question. We could have also used completing the square. We want to complete the square on the quadratic negative two π₯ squared plus 12π₯ minus 25. Weβll start by taking the constant factor of negative two outside of our first two terms. This gives us negative two multiplied by π₯ squared minus six π₯ minus 25.

Next, we want to write π₯ squared minus six π₯ in the form π₯ plus π all squared. And to do this, we recall we need to halve the coefficient of π₯. Therefore, if we distribute the exponent over π₯ minus three all squared, we have π₯ minus three all squared is π₯ squared minus six π₯ plus nine. We can then subtract nine from both sides of the equation to get an expression for π₯ squared minus six π₯. We have π₯ squared minus six π₯ is π₯ minus three all squared minus nine. We can then substitute this into our expression. This gives us negative two multiplied by π₯ minus three all squared minus nine minus 25.

If we distribute the factor of negative two over our parentheses, we get negative two times π₯ minus three all squared plus 18 minus 25. And we can simplify this. 18 minus 25 is equal to negative seven. And therefore, by completing the square, weβve rewritten our function as negative two times π₯ minus three all squared minus seven. And we can use this to find the turning point of our parabola. But letβs start by recalling why this is true. We see that the expression π₯ minus three is squared. And the square of a number is always greater than or equal to zero.

But then, we multiply this number by negative two, which means the entire first term of this expression is always less than or equal to zero. And this means to make this value as large as possible, we should try and make our first term equal to zero, since this is the biggest our first time can be. And this happens when π₯ is equal to three. Therefore, π₯ is equal to three is where the maximum value of this quadratic occurs. We can substitute it into our function and we will also get negative seven. And of course, since this is the value where our first term is equal to zero, the maximum value of our function is the second term, the constant value of negative seven.

Therefore, weβve seen two different ways of finding the maximum value of the quadratic function π¦ is equal to negative two π₯ squared plus 12π₯ minus 25. In both cases, this value was negative seven.