Answer: 200
The quadratic function f(x) = a(x - h)2 + k, a not equal to zero, is said to be in standard form. If a is positive, the graph opens upward, and if a is negative, then it opens downward. The line of symmetry is the vertical line x = h, and the vertex is the point (h,k).
Any quadratic function can be rewritten in standard form by completing the square. (See the section on solving equations algebraically to review completing the square.) The steps that we use in this section for completing the square will look a little different, because our chief goal here is not solving an equation.
Note that when a quadratic function is in standard form it is also easy to find its zeros by the square root principle.
Example 3.
Write the function f(x) = x2 - 6x + 7 in standard form. Sketch the graph of f and find its zeros and vertex.
f(x) = x2 - 6x + 7.
= (x2 - 6x )+ 7. Group the x2 and x terms and then complete the square on these terms.
= (x2 - 6x + 9 - 9) + 7.
We need to add 9 because it is the square of one half the coefficient of x, (-6/2)2 = 9. When we were solving an equation we simply added 9 to both sides of the equation. In this setting we add and subtract 9 so that we do not change the function.
= (x2 - 6x + 9) - 9 + 7. We see that x2 - 6x + 9 is a perfect square, namely (x - 3)2.
f(x) = (x - 3)2 - 2. This is standard form.
From this result, one easily finds the vertex of the graph of f is (3, -2).
To find the zeros of f, we set f equal to 0 and solve for x.
(x - 3)2 - 2 = 0.
(x - 3)2 = 2.
(x - 3) = ± sqrt(2).
x = 3 ± sqrt(2).
To sketch the graph of f we shift the graph of y = x2 three units to the right and two units down.
If the coefficient of x2 is not 1, then we must factor this coefficient from the x2 and x terms before proceeding.
Example 4.
Write f(x) = -2x2 + 2x + 3 in standard form and find the vertex of the graph of f.
f(x) = -2x2 + 2x + 3.
= (-2x2 + 2x) + 3.
= -2(x2 - x) + 3.
= -2(x2 - x + 1/4 - 1/4) + 3.
We add and subtract 1/4, because (-1/2)2 = 1/4, and -1 is the coefficient of x.
= -2(x2 - x + 1/4) -2(-1/4) + 3.
Note that everything in the parentheses is multiplied by -2, so when we remove -1/4 from the parentheses, we must multiply it by -2.
= -2(x - 1/2)2 + 1/2 + 3.
= -2(x - 1/2)2 + 7/2.
The vertex is the point (1/2, 7/2). Since the graph opens downward (-2 < 0), the vertex is the highest point on the graph.
Exercise 2:
Write f(x) = 3x2 + 12x + 8 in standard form. Sketch the graph of f ,find its vertex, and find the zeros of f. Answer
Alternate method of finding the vertex
In some cases completing the square is not the easiest way to find the vertex of a parabola. If the graph of a quadratic function has two x-intercepts, then the line of symmetry is the vertical line through the midpoint of the x-intercepts.
The x-intercepts of the graph above are at -5 and 3. The line of symmetry goes through -1, which is the average of -5 and 3. (-5 + 3)/2 = -2/2 = -1. Once we know that the line of symmetry is x = -1, then we know the first coordinate of the vertex is -1. The second coordinate of the vertex can be found by evaluating the function at x = -1.
Example 5.
Find the vertex of the graph of f(x) = (x + 9)(x - 5).
Since the formula for f is factored, it is easy to find the zeros: -9 and 5.
The average of the zeros is (-9 + 5)/2 = -4/2 = -2. So, the line of symmetry is x = -2 and the first coordinate of the vertex is -2.
The second coordinate of the vertex is f(-2) = (-2 + 9)(-2 - 5) = 7*(-7) = -49.
Therefore, the vertex of the graph of f is (-2, -49).
. Set the derivative equal to 0 and factor to find the critical numbers. 
, so x = -3 and x = 1/3. We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing. From there we will look at our points to determine which fall into the "decreasing" category. Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞. In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval. In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval. In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval. In order to determine where our points in question fall, look to the x value. The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12). The point (-3, -18) is already a min value.