Answer :
Explanation
In this problem, we have to:
• a) determine the number of intersections between two functions,
,• b) find the points of intersection.
The discriminant of a polynomial y = ax² + bx + c is given by:
[tex]\Delta=b^2-4ac.[/tex]According to the value of the discriminant, we have:
• for Δ > 0 ⇒ two solutions,
,• for Δ = 0 ⇒ one solution,
,• for Δ < 0 ⇒ no solutions.
The solutions for y = 0 are given by:
[tex]x=\frac{-b\pm\sqrt{\Delta}}{2a}.[/tex]ii) We have the functions:
[tex]\begin{gathered} f(x)=5x^2-x+1, \\ g(x)=2x+3. \end{gathered}[/tex](a) At the intersection of the functions, we have:
[tex]\begin{gathered} f(x)=g(x), \\ 5x^2-x+1=2x+3, \\ 5x^2-3x-2=0. \end{gathered}[/tex]Comparing this polynomial with the general equation from above, we see that:
• a = 5,
,• b = -3,
,• c = -2.
To find the number of intersections, we compute the discriminant of this polynomial:
[tex]\Delta=(-3)^2-4\cdot5\cdot(-2)=9+40=49>0\Rightarrow\text{ two solutions.}[/tex]So we have two intersections.
(b) The values of x at the intersections are:
[tex]x=\frac{-(-3)\pm\sqrt{49}}{2\cdot5}=\frac{3\pm7}{10}\Rightarrow\begin{cases}x_1=\text{ }{1} \\ x_2={-\frac{2}{5}}.\end{cases}[/tex]Replacing these values in the equation of one of the polynomials, we get:
[tex]\begin{gathered} (x_1,y_1)=(1,2\cdot1+3)=(1,5), \\ (x_2,y_2)=(-\frac{2}{5},2\cdot(-\frac{2}{5})+3)=(-\frac{2}{5},\frac{11}{5})=(-0.4,2.2). \end{gathered}[/tex]Answer(ii)
• (a) There are 2 intersections
,• (b) The intersections are (1,5) and (-0.4, 2.2)