<p>The</a> Official SAT Question of the Day</p>
<p>y = -2x^2 + bx +5</p>
<p>In the xy-plane, the graph of the equation above assumes its maximum value at . What is the value of b?</p>
<p>Hi! the answer is 8, this is how I did it:</p>
<p>The equation is y = -2x^2 + bx + 5
For a maximum value, dy/dx (which is what you get when you differentiate the equation) equals 0
When you differentiate the equation, you get dy/dx = -4x + b
For the maximum value, x=2 and dy/dx = 0 → so 0 = -4(2) + b → so b = 8</p>
<p>I solved it the same way as gabby, but if you don’t know Calculus yet you can solve it with algebra as well.
Because that equation is of a downward opening parabola, the maximum value is going to be at the vertex. Knowing the X value=2 at this point you can rewrite the equation as:
y=-2(x-2)^2+k
-2 is the slope (which you get from the squared term in the original equation)
(x-2) was determined because we know the x-value at the vertex is 2 (remember that numbers inside parenthesis are opposite signs for these equations)
and the plus k is what you are solving for.</p>
<p>Go ahead and foil out that -2(x-2)^2 to get -2x^2+8x-8+k. Now that you have the equation back in its original form, remember that ‘b’ was the coefficient for x. Identify the x term as 8x, and now you know that b=8!</p>
<p>The SAT will never require calculus nor the algebraic set-up and manipulation offered above and in the official explanation (which frankly doesn’t even make sense to me).</p>
<p>When you see a question like this and the answer is not immediately obvious, draw the graph. It will become obvious that the formula represents a parabola that opens downward, crosses the y-axis at +5, and has some unknown maximum value when x=2 (the vertex of the parabola).</p>
<p>Look at the picture you have drawn and it also becomes obvious that you know one other point on the parabola. Parabolas are symmetrical. If y=5 when x=0, and the vertex is at x=2, then y must also equal 5 when x equals what?</p>
<p>If y=5 two places to the left of the vertex then it must also =5 two places to the right of the vertex. When x=4, y must =5 also.</p>
<p>Plug those numbers into the original equation and solve for b. The given formula does not have to be manipulated in any way.</p>
<p>I think this is a much more intuitive solution that doesn’t require much more than logic and very basic ‘solve for one unknown’ algebra.</p>
<p>I just used the equation x= -b/2a
x is given, which is 2. a is -2 because it is the first leading coefficient. Plug those values in and solve for b. </p>
<p>Sent from my SPH-D710 using CC</p>
<p>I was wondering the same thing! Thanks for posting this question. Great explanation</p>
<p>-b/2a</p>
<p>-b/2(-2)</p>
<p>b/4=2</p>
<p>b=8</p>
<p>-b/2a is the formula for finding the x value of the vertex of the quadratic equation, if f(x)=ax^2+bx+c</p>
<p>Yeah, all you need to remember is that the vertex of a quadratic polynomial in the form y = ax^2 + bx + c is at x = -b/2a. It’s very easily proven using calculus (since setting the derivative to zero yields 2ax + b = 0 → x = -b/2a) but it can also be proven with algebra.</p>
<p>Note that ax^2 + bx + c = a(x - b/2a)^2 + (c - (b/2a)^2). Ignore the constant term for now so we are left with a(x - b/2a)^2. Since squares are always non-negative, the minimum value of (x - b/2a)^2 occurs at x = -b/2a, which results in a max. or min. for y (depending on the sign of a).</p>
<p>This only works for quadratics – for cubics, you would likely need calculus.</p>