-4y2+16y-x-17=0 is a parabola lying on its side. We know this because it's y, rather than x, that's squared while x appears by itself.
The general form for such a parabola is x=(y-k)2+h where (h,k) is the vertex point.
x=-4y2+16y-17,
x=-4(y2-4y)-17,
x=-4(y2-4y+4-4)-17,
x=-4[(y-2)2-4]-17,
x=-4(y-2)2+16-17,
x=-4(y-2)2-1 tells that the vertex is at (-1,2). -4 is a negative coefficient which indicates that whether y increases or decreases, x is always negative and the maximum value of x (=-1) is when y=2, that is, the vertex, and the parabola lies to the left of its vertex.
When y=0, x=-17, the x intercept. The vertex lies on the axis of symmetry which is horizontal at y=2. These are important details that help to draw the graph. The directrix and focus do not really assist in the drawing.
I find it useful to draw a horizontal line which is the same distance above the axis of symmetry (green on the picture) as the axis of symmetry is above the x-axis. In this case, the line is y=4 (also shown in green on the picture) because the axis of symmetry is y=2. So mark the point (-17,2) on the other side of the x-intercept. This helps to sketch the parabola (in red) which is symmetrical about the axis of symmetry. To summarise, the parabola lies on its side with its arms pointing to the left. Mark the vertex (-1,2), the x-intercept (-17,0) the reflection of the x-intercept (-17,4). These points should guide you to sketch the parabola fairly accurately.
