finding minimum/maximum in quadratics

An SAT Math micro-topic under Quadratic and exponential word problems (Advanced Math). Free to read — no account needed.

In vertex form a(x−h)2+ka(x - h)^2 + k, a quadratic's extreme value is easy to read off. The squared part (x−h)2(x - h)^2 is always ≥0\ge 0, so it is smallest (zero) exactly at x=hx = h, which is where the maximum or minimum occurs.
quadratic_min_max.png
The figure shows both cases. When a>0a > 0 the parabola opens upward and the vertex is the lowest point, so kk is the minimum. When a<0a < 0 it opens downward and the vertex is the highest point, so kk is the maximum.
Here is why: since (x−h)2≥0(x - h)^2 \ge 0, multiplying by a positive aa keeps it ≥0\ge 0, so a(x−h)2+k≥ka(x - h)^2 + k \ge k (minimum kk); multiplying by a negative aa makes it ≤0\le 0, so a(x−h)2+k≤ka(x - h)^2 + k \le k (maximum kk).
If the quadratic is expanded instead, the sign of the x2x^2 coefficient still tells the direction: positive opens up (a minimum), negative opens down (a maximum). For −12x2+8-\frac{1}{2}x^2 + 8, the coefficient is negative, so 88 is the maximum.

Worked examples

What is the minimum value of (x−2)2+5(x - 2)^2 + 5?
Since (x−2)2≥0(x - 2)^2 \ge 0, the smallest it can be is 00, at x=2x = 2.
So the minimum value is 0+5=50 + 5 = 5.
What is the maximum value of R=−40(h−3)2+80R = -40(h - 3)^2 + 80?
Since (h−3)2≥0(h - 3)^2 \ge 0 and it is multiplied by −40-40, the term −40(h−3)2≤0-40(h - 3)^2 \le 0.
So RR is largest when that term is 00, at h=3h = 3, giving a maximum of 8080.
A fountain's height is w(x)=−12(x2−16)w(x) = -\frac{1}{2}(x^2 - 16). What is its maximum height?
Expand: w(x)=−12x2+8w(x) = -\frac{1}{2}x^2 + 8.
The x2x^2 coefficient is negative, so the maximum is the constant 88, reached at x=0x = 0.

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