-gre.ge.04 intersection of parabola and line

[karush]

$\textbf{xy-plane}$ above shows one of the two points of intersection of the graphs of a linear function and and quadratic function.
The shown point of intersection has coordinates $\textbf{(v,w)}$ If the vertex of the graph of the quadratic function is at $\textbf{(4,19)}$,
what is the value of $\textbf{v}$?
${-6}\quad {6}\quad {5}\quad {7}\quad {8}$

ok before I plow into this one it seems obvious that v could not be known for certain by observation
(the graph does not look it is to scale)
so then we can only proceed with the intersections of the equations of
$$y=a(x-4)^2 +19 \quad y=\dfrac{9}{2}x-9$$

unless some other quickie could apply

 

[skeeter]

the line has equation $y=4x-9$

note the secant line from (0,3) to (4,19) is parallel to the line $y=4x-9$

 

[karush]

the line has equation $y=4x-9$
note the secant line from (0,3) to (4,19) is parallel to the line $y=4x-9$

$-\left(x^{2}-8x+16\right)-4x+9+19=0$
$x^{2}-4x-12=0$
$(x-6)(x+2)$
v=6

ok I couldn't see how the secant would make things obvious

 

[HOI]

View attachment 10810​

$\textbf{xy-plane}$ above shows one of the two points of intersection of the graphs of a linear function and and quadratic function.
The shown point of intersection has coordinates $\textbf{(v,w)}$ If the vertex of the graph of the quadratic function is at $\textbf{(4,19)}$,
what is the value of $\textbf{v}$?
${-6}\quad {6}\quad {5}\quad {7}\quad {8}$

ok before I plow into this one it seems obvious that v could not be known for certain by observation
(the graph does not look it is to scale)
so then we can only proceed with the intersections of the equations of
$$y=a(x-4)^2 +19 \quad y=\dfrac{9}{2}x-9$$

No! If x= 2, this gives y= 9- 9= 0, not -1. The point (2, -1) is just below the x-axis, not on it.

unless some other quickie could apply

 

[skeeter]

$-\left(x^{2}-8x+16\right)-4x+9+19=0$
$x^{2}-4x-12=0$
$(x-6)(x+2)$
v=6

ok I couldn't see how the secant would make things obvious

$\dfrac{w - (-1)}{v - 2} = 4$

note from the graph that $4 <v < 8$ and $3 < w < 19$

so, only two possible coordinates for $(v,w)$ ...

$(5,11)$ and $(6, 15)$

$(5,11)$ would be vertically midway between $(0,3)$ and $(4,19)$ if it were $(v,w)$.

 

[karush]

ok i see
mahalo much