Is it possible for a system of two Leonard places to have no solution explain your reasoning
1 answer:
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Answer:
Look to the attached graph
Step-by-step explanation:
* Lets revise how to graph the quadratic function
- Find the vertex of it
- Find the y-intercept
- Find the x-intercept
∵ f(x) = (x - 5)² + 1 ⇒ the completing square form
- The completing square form for any quadratic is
( x - h)² + k, where h and k are the coordinate of the vertex point
* Lets compare the two forms
∵ (x - h)² + k = (x - 5)² + 1
∴ h = 5 and k = 1
∴ The vertex of the parabola is (5 , 1)
- To find the x-intercept put f(x) = 0
∵ (x - 5)² + 1 = 0 ⇒ subtract 1 from both sides
∴ (x - 5)² = -1 ⇒ the square can not give -ve number
∴ The parabola does not intersect the x-axis
- To find the y-intercept put x = 0
∵ f(0) = (0 - 5)² + 1 = 25 + 1 = 26
∴ The parabola intersects the y-axis at point (0 , 26)
- The parabola is opened upward because the coefficient of x² is +ve
* Now lets graph it
- Look to the attached graph
