# Question Video: Finding the Local Maximum and Minimum Values of a Quadratic Function Mathematics

Determine the local maximum and minimum values of the function 𝑦 = −3𝑥² − 6𝑥 − 4.

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### Video Transcript

Determine the local maximum and minimum values of the function 𝑦 equals negative three 𝑥 squared minus six 𝑥 minus four.

First, we recall that at critical points, the first derivative of the function — in this case d𝑦 by d𝑥 — is equal to zero. So our first step is going to be to find the first derivative of this function. By applying the power rule of differentiation, we find that d𝑦 by d𝑥 is equal to negative six 𝑥 minus six. We then set our expression for d𝑦 by d𝑥 equal to zero and solve for 𝑥, giving 𝑥 is equal to negative one. Our function, therefore, has one critical point, which occurs when 𝑥 is equal to negative one.

Next, we need to evaluate the function at the critical point, which we do by substituting 𝑥 equals negative one into the equation we’ve been given. We obtain 𝑦 equals negative three multiplied by negative one squared minus six multiplied by negative one minus four which simplifies to negative one. This tells us then that the only critical point of this function is the point with coordinates negative one, negative one. But we need to determine whether this is a local minimum or local maximum, which we’ll do by applying the second derivative test.

To find the second derivative, we need to differentiate our first derivative with respect to 𝑥. So we’re finding the derivative of negative six 𝑥 minus six with respect to 𝑥. Applying the power rule, we see that this derivative is just equal to negative six. Now, this second derivative is actually just a constant because we’ve differentiated a quadratic expression twice. So we don’t need to substitute the 𝑥-coordinate at our critical point in in order to evaluate because the second derivative is constant for all values of 𝑥. We note that negative six is less than zero. We recall that if the second derivative of a function is negative at the critical point, then the critical point is a local maximum. So the point negative one, negative one is indeed a local maximum of this function.

So we can conclude that this function has no local minimum value but has a local maximum value of negative one. Notice that is the value of the function itself that we are giving here, not the 𝑥-value, although they are both the same in this instance. We can also confirm this result using our knowledge of the graphs of quadratic functions. As the coefficient of 𝑥 squared in this curve is negative, the graph of this quadratic will be a negative parabola. We know that parabolas have only a single critical point. And if the coefficient of 𝑥 squared is negative, then their critical point will be a local maximum.