lesson
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If a stunt plane flies in a straight line while a firework arcs through the sky along a curve, how can you pinpoint the exact moments their flight paths cross?
In math, this scenario is a linear-quadratic system, made of one linear equation (y=mx+b) and one quadratic equation (y=ax2+bx+c).
Before we start drawing on the coordinate plane, how many times can a straight line and a curved parabola actually cross?
Possible Number of Solutions
A line can slice straight through a parabola at two points, graze it at exactly one point (as a tangent line), or miss it entirely with no real solutions.
📊A clean 3-panel comparison diagram on a light gray background (#f8f9fa). Panel 1 (left): 'Two Solutions (Secant)' shows a blue parabola y = x^2 - 2 and an orange line cutting across it, with two bright red dots at the intersection points labeled '(x_1, y_1)' and '(x_2, y_2)'. Panel 2 (middle): 'One Solution (Tangent)' shows a blue parabola with an orange horizontal line barely touching its vertex at one red dot labeled '(x_0, y_0)'. Panel 3 (right): 'No Solution' shows a blue parabola opening upward and an orange line passing completely below it with a label 'No Intersection'. All panels use neat coordinate grids with text color #1e2945 and borders #e6e6e6.
Let's put this into practice with a concrete system and see how to locate the intersection points step by step.
Graphing the System
Consider the system with quadratic equation y=x2−4x+3 and linear equation y=−x+1.
To graph the parabola, find its turning point or vertex using the formula x=−2ab, which comes from the line of symmetry halfway between its roots.
For y=x2−4x+3, a=1 and b=−4, giving x=−2(1)−4=2. Plugging in x=2 gives y=(2)2−4(2)+3=−1, so the vertex sits at (2,−1).
Plot the vertex (2,−1) along with symmetric points like (0,3), (1,0), (3,0), and (4,3) to draw the smooth curve.
Next, graph the line y=−x+1 by placing a point at the y-intercept (0,1) and using the slope m=−1 (down 1, right 1).