\(f(x) = ax^2 + 4x + c\)
In the given quadratic function, \(a\) and \(c\) are constants. The graph of \(y=f(x)\) in the xy-plane is a parabola that opens upward and has a vertex at the point \((h, k)\), where \(h\) and \(k\) are constants. If \(k<0\) and \(f(-9) = f(3)\), which of the following must be true?
We can start off with \(f(-9)\) being equal to \(f(3)\).
\(f(-9)\) comes to \(81a-36+c\) and \(f(3)\) comes to \(9a+12+c\). If we equate both of these, we can solve for \(a\).
\(81a-36+c=9a+12+c\)
We can DESTROY the \(c\) on both sides.
\(81a-36=9a+12\)
Solving for \(a\), we get:
\(a=\frac{2}{3}\)
Now we know that condition 2 is false.
Now, we will simplify our expression for \(f(x)\) by replacing the \(a\) with \(\frac{2}{3}\).
\(f(x) = \frac{2}{3}x^2 + 4x + c\)
We are also given the coordinates of the vertex of this parabola: \(h,k\). If we were to complete the square on our brand new \(f(x)\), we might be able to obtain \(h\) or \(k\).
We'll do just that: we have \(f(x)=\frac{2}{3}(x+3)^2-6+c\). Since we have some extra information about \(k\), we can equate it to \(-6+c\).
\(-6+c=k\). From here, we can actually solve for \(c\) to verify condition 1. \(c=k+6\).
Let's test this two ways. We'll asign \(k\) any two values less than zero (remember \(k<0\)): \(-2\) and \(-10\). If we plug either value into our equation in \(c\), we can see if the first condition holds.
For \(k=-2\), \(c\) comes to \(4\). Positive \(4\). However, for \(k=-10\), \(c\) comes to negative \(4\). So condition 1 isn't always true.
Hence, both conditions are false.
This article was written on 2024-07-26. If you have any thoughts, feel free to send me an email with them. Have a nice day!