<p>Hi guys please help me with this question:
y=- 2(x-2)+3
in tje xy plane, line L passes through the point (4,-5) and the vertex of the parabola with the equation above , what is the slope of line L ?
(a)-4
(b)-1/4
(c) 0
(d) 1/4
(e) 4
how can I get the vertex and the slope :) </p>
<p>Do you mean y = -2(x-2)^2 + 3? What you typed is a linear equation.</p>
<p>The vertex of the graph of y = -2(x-2)^2 + 3 is at (2,3). The slope of line L connecting (4,-5) and (2,3) is (-5-3)/(4-2) = -4.</p>
<p>How did you get (2,3)?</p>
<p>In general, a parabola of the form y = a(x-b)^2 + c (a,b,c reals, a!=0) has a vertex at (b,c). This is because x=b minimizes the value of (x-b)^2, which minimizes or maximizes y (depending on the sign of a).</p>
<p>Thank you a lot 
</p>