How do you graph #y = x^2 + 7x - 1#?

1 Answer

Answer:

The given equation can be changed into a standard equation of parabola: #(x-x_1)^2=4a(y-y_1)# having vertex at #(x_1, y_1)# & focus at #(x_1, y_1+a)#

Explanation:

Given equation: #y=x^2+7x-1#
#y=x^2+2\cdot \frac{7}{2}\cdotx+(\frac{7}{2})^2-(\frac{7}{2})^2-1#
#y=(x+\frac{7}{2})^2-\frac{51}{2}#
#(x+\frac{7}{2})^2=y+\frac{51}{2}#
#(x+\frac{7}{2})^2=4\cdot \frac{1}{4}(y+\frac{51}{2})#
comparing the above equation with #(x-x_1)^2=4a(y-y_1)#
we get #x_1=-7/2, y_1-51/2, a=1/4#
then the given parabola has vertex at #(x_1, y_1)\equiv(-7/2, -51/2)# & focus at #(x_1, y_1+a)\equiv(-7/2,-101/4 )#
Points of intersection of parabola: #y=x^2+7x-1# with x-axis where #y=0# are #(\frac{-7+\sqrt{53}}{2}, 0)# & #(\frac{-7-\sqrt{53}}{2}, 0)# and the point of intersection with y-axis is #(0, -1)#
Now, locate the vertex & locus & draw the axis of parabola. Draw free hand curve passing through above points of intersection with the coordinate axes