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

Question

Eli Assouline · Accepted 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)# 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

Hazel Anglin · Answer

undefined To graph the equation $ y = x^2 + 7x - 1 $, you can follow these steps:

1. Identify the vertex using the formula $ x = -\frac{b}{2a} $, where $ a = 1 $ (coefficient of $ x^2 $) and $ b = 7 $ (coefficient of $ x $).
2. Substitute the value of $ x $ obtained from step 1 into the equation to find the $ y $-coordinate of the vertex.
3. Plot the vertex on the coordinate plane.
4. Determine additional points by choosing values of $ x $ and finding the corresponding $ y $-values using the equation.
5. Plot these points and draw a smooth curve through them to complete the graph.