# Math Help - Parabol Questions

1. ## Parabol Questions

Question1: $g(x)=x^2+5x+3+m$ is the equation of the symmetry with respect to y axis of $f(x)=x^2-5x+4$ ,find m
Question2: If f:RtoR and $f(x)=x^2-(3k-5)x+m+4$ are given. Find the area of the triangle which is bounded by the symmetry axis of $f(x)$,the line $y=-x$ and the $Ox$ axis. Thanks in advance for the solution of these questions.

2. Originally Posted by JohnDoe
Question1: $g(x)=x^2+5x+3+m$ is the equation of the symmetry with respect to y axis of $f(x)=x^2-5x+4$ ,find m
Question2: If f:RtoR and $f(x)=x^2-(3k-5)x+m+4$ are given. Find the area of the triangle which is bounded by the symmetry axis of $f(x)$,the line $y=-x$ and the $Ox$ axis. Thanks in advance for the solution of these questions.
Q1: The y-intercept of f is a fixed point of the reflection about the y-axis. Therefore the point (0, 4) belongs to the graph of g too:

$g(0)=3+m = 4 ~\implies~m = 1$

Q2: Do you use the value of m from Q1 in this question too? And what does this mean?: $Ox$

3. Q2 doesn't have any relation with Q1 they are totally apart and i think Ox is y=0 line.

4. Originally Posted by JohnDoe
[snip]
Question2: If f:RtoR and $f(x)=x^2-(3k-5)x+m+4$ are given. Find the area of the triangle which is bounded by the symmetry axis of $f(x)$,the line $y=-x$ and the $Ox$ axis. Thanks in advance for the solution of these questions.
There are a number of ways of showing that

the symmetry axis is the vertical line $x = \frac{3k-5}{2}$.

Each way finds the x-coordinate of the turning point of the given parabola.

The area of a triangle is $\frac{1}{2} \text{ base} \times \text{height}$.

A simple sketch graph shows that:

The base of the triangle is $\left| \frac{3k - 5}{2} \right|$.

The height of the triangle is $\left| -\left(\frac{3k - 5}{2} \right) \right| = \left| \frac{5 - 3k}{2} \right|$.

You need magnitudes because $\frac{3k - 5}{2}$ can be positive or negative depending on the value of k ......