Math!!!

<ol>
<li>The graph in the xy-plane of the quadratic function f contains the points(0,0) (1,5) and (5,5). What is the maximum value of f(x)?</li>
</ol>

<p>and 18. In the rectangle above, the ratio of AB to Bc is 1:1, the ratio of CD to DE is 1:3 and the ratio of FG to GA is 1:3, what fraction of the area of the rectangle is shaded?</p>

<p>I just assumed that the figure wasn’t drawn to scale even though it actually is and assumed that CE was 4 and AC was 2 which means the total area is 8. Then i subtracted the areas of the triangles that arn’t shaded. Will this technique always work or is it just this one time? and is this the most efficient/accurate way to solve this?
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<ol>
<li>A quadratic function is of the form y=ax^2 + bx + c. If you plug in the 3 pairs of values in, you get three equations, and you can solve for a, b, and c. </li>
</ol>

<p>The maximum value is at the discriminant -b/2a, which should be equal to 3, since you get the same value at 1 and 5, and a parabola is symmetrical. Put that in your function.</p>

<ol>
<li>Yeah, this technique works. It’s all in proportions, so you can assign whatever values you want and the fractions will come out right. In this case, your way is the definitely the quickest.</li>
</ol>

<p>Thanks conquerer, does this technique always work for problems like these? even if i am distorting the values somewhat since my width is longer than my length lol.</p>

<p>and will the actual SAT ever test something like question 19? this is from a prep company.</p>

<p>19 seems quite a bit harder than anything I’ve seen on the real thing, but that was back in 7th grade, so I really dunno.</p>

<p>And yeah, it should work for all problems like that, and you can put whatever lengths you want as long as every ratio is satisfied. Though depending on the question, it might be better to add areas together rather than starting with something and subtracting.</p>

<p>and another.
16. If a,b,m,n, and p are all positive integers greater than 1, and if (ab)^m=(a^n)(b^p) which of the following are true?</p>

<p>I.2m=n+p
II.a=b
III.if n =p then n=m</p>

<p>III is true: then the right hand side is equal to (a^n)(b^n) = (ab)^n so n=m.</p>

<p>II is false: there’s no reason a should have to equal b. It’s especially clear if you let n=p: a and b can be anything.</p>

<p>I is false, but it will be true if a and b are relatively prime; it’s pretty tricky. But try a = 2, b = 4, m = 3, n = 7, p = 1. Then I is clearly not true.</p>