Answer:
[tex]\sf a) \quad P=(-2, 2) \: \textsf{ and }\: \sf Q= \left(\dfrac{3}{4}, \dfrac{27}{8}\right)[/tex]
[tex]\textsf{b)} \quad x\leq -2 \:\:\: \textsf {or }\:\:x\geq \dfrac{3}{4}[/tex]
Step-by-step explanation:
Part (a)
Given equations:
[tex]\begin{cases}y=\dfrac{1}{2}x+3\\\\y=2x^2+3x\end{cases}[/tex]
Points P and Q are the points of intersection of the given equations.
Therefore, to find the points of intersection, equate the equations and solve for x:
[tex]\implies 2x^2+3x=\dfrac{1}{2}x+3[/tex]
[tex]\implies 2x^2+\dfrac{5}{2}x-3=0[/tex]
[tex]\implies 2\left(2x^2+\dfrac{5}{2}x-3\right)=2(0)[/tex]
[tex]\implies 4x^2+5x-6=0[/tex]
[tex]\implies 4x^2+8x-3x-6=0[/tex]
[tex]\implies 4x(x+2)-3(x+2)=0[/tex]
[tex]\implies (4x-3)(x+2)=0[/tex]
Apply the zero-product property:
[tex]\implies 4x-3=0 \implies x=\dfrac{3}{4}[/tex]
[tex]\implies x+2=0 \implies x=-2[/tex]
Substitute the found values of x into one of the equations to find the y-coordinates of points P and Q:
[tex]x_P=-2 \implies y_P=\dfrac{1}{2}(-2)+3=2[/tex]
[tex]x_Q=\dfrac{3}{4} \implies y_Q=\dfrac{1}{2}\left(\dfrac{3}{4}\right)+3=\dfrac{27}{8}[/tex]
Therefore, the coordinates of P and Q are:
[tex]\sf P=(-2, 2) \: \textsf{ and }\: \sf Q= \left(\dfrac{3}{4}, \dfrac{27}{8}\right)[/tex]
Part (b)
Given inequality:
[tex]2x^2+3x \geq \dfrac{1}{2}x+3[/tex]
The range of values that satisfies the given inequality is the range of x-values where the curve is equal to or higher than the line.
The curve is equal to or higher than the line for x-values equal to or less than point P and x-values equal to or more than point Q:
[tex]x\leq -2 \:\:\: \textsf {or }\:\:x\geq \dfrac{3}{4}[/tex]
