To graph the parabola, we first need to identify a, b, and c.
2x^2-3x-y=0 ⇕ y= 2x^2+( - 3)x+ 0
For this equation we have that a= 2, b= - 3, and c= 0. Now, we can find the vertex using its formula. To do this, we will need to think of y as a function of x, y=f(x).
Vertex of a Parabola: ( - b/2 a,f(- b/2 a) )
Let's find the x-coordinate of the vertex.
The y-coordinate of the vertex is - 98. Therefore, the vertex is at the point ( 34,- 98). With this, we also know that the axis of symmetry of the parabola is the line x= 34. Next, let's find two more points on the curve, one on each side of the axis of symmetry.
x
2x^2-3x
y=2x^2-3x
0
2( 0)^2-3( 0)
0
2
2( 2)^2-3( 2)
2
Both ( 0, 0) and ( 2, 2) are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.
Graphing the Line
Let's now graph the linear function on the same coordinate plane. We will start by rewriting the given equation in slope-intercept form.
We can now identify the slope m and y-intercept b of the given function.
y=5/2x-9/4 ⇔ y= 5/2x+( - 9/4)
The slope of the line is 52 and the y-intercept is - 94.
Finding the Solutions
Finally, let's try to identify the coordinates of the points of intersection of the parabola and the line.
It looks like the points of intersection occur at ( 12,- 1) and (2 14, 3 38), which is equivalent to ( 94, 278).
Checking the Answer
To check our answers, we will substitute the values of the points of intersection in both equations of the system. If they produce true statements, our solution is correct. Let's start with ( 12, - 1).
