To graph the parabola, we first need to identify a, b, and c.
f(x)=x^2-3x-10 ⇔ f(x)= 1x^2+( -3)x+( -10)
For this equation we have that a= 1, b= -3, and c= -10. Now, we can find the vertex using its formula.
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 - 494. Thus, the vertex is at the point ( 32, 494). With this information, we also know that the axis of symmetry of the parabola is the line x= 32. Next, let's find two more points on the curve, one on each side of the axis of symmetry.
x
x^2-3x-10
f(x)=x^2-3x-10
^2-3( )-10
-10
3
3^2-3(3)-10
-10
Both ( ,-10) and (3,-10) 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.
For a linear equation written in slope-intercept form, we can identify its slope m and y-intercept b.
g(x)=-5x-7 ⇔ g(x)=-5x+( -7)
The slope of the line is -5 and the y-intercept is -7.
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 (-3,8) and (1,-12).
