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Q(x)=(P(2x))/(2)
n the equation shown above, 
Q and 
P are functions. If 
(x_(0),y_(0)) is a point on the graph of 
y=Q(x), which of the following is a point on the graph of 
y=P(x) ?

{:[(x_(0),y_(0))],[(2x_(0),(y_(0))/(2))],[((x_(0))/(2),2y_(0))],[(2x_(0),2y_(0))]:}

Q(x)=P(2x)2 Q(x)=\frac{P(2 x)}{2} \newlinen the equation shown above, Q Q and P P are functions. If (x0,y0) \left(x_{0}, y_{0}\right) is a point on the graph of y=Q(x) y=Q(x) , which of the following is a point on the graph of y=P(x) y=P(x) ?\newline(x0,y0)(2x0,y02)(x02,2y0)(2x0,2y0) \begin{array}{l} \left(x_{0}, y_{0}\right) \\ \left(2 x_{0}, \frac{y_{0}}{2}\right) \\ \left(\frac{x_{0}}{2}, 2 y_{0}\right) \\ \left(2 x_{0}, 2 y_{0}\right) \end{array}

Full solution

Q. Q(x)=P(2x)2 Q(x)=\frac{P(2 x)}{2} \newlinen the equation shown above, Q Q and P P are functions. If (x0,y0) \left(x_{0}, y_{0}\right) is a point on the graph of y=Q(x) y=Q(x) , which of the following is a point on the graph of y=P(x) y=P(x) ?\newline(x0,y0)(2x0,y02)(x02,2y0)(2x0,2y0) \begin{array}{l} \left(x_{0}, y_{0}\right) \\ \left(2 x_{0}, \frac{y_{0}}{2}\right) \\ \left(\frac{x_{0}}{2}, 2 y_{0}\right) \\ \left(2 x_{0}, 2 y_{0}\right) \end{array}
  1. Calculate y-value on y=P(x)y=P(x): Since y0=Q(x0)y_{0}=Q(x_{0}), we have y0=P(2x0)2y_{0}=\frac{P(2x_{0})}{2}. To find the corresponding y-value on y=P(x)y=P(x), we need to find P(2x0)P(2x_{0}).
  2. Multiply by 22: We can multiply both sides of y0=P(2x0)2y_{0}=\frac{P(2x_{0})}{2} by 22 to get 2y0=P(2x0)2y_{0}=P(2x_{0}). This gives us the y-value of the point on y=P(x)y=P(x) when xx is 2x02x_{0}.
  3. Find corresponding point: Therefore, the corresponding point on the graph of y=P(x)y=P(x) is (2x0,2y0)(2x_{0}, 2y_{0}).

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