Answer:
See bolded below.
Step-by-step explanation:
As you said, you only need help with no. 2:
a) Check the attachments for the lines
b) The equations are, respectively:
y = -4x + 1 [intersects parabola at 1 point]
Check:
[tex]\begin{bmatrix}y=\left(x-2\right)^2-3\\ y=-4x+1\end{bmatrix}[/tex]
[tex]\begin{bmatrix}-4x+1=\left(x-2\right)^2-3\end{bmatrix}[/tex]
[tex]\begin{bmatrix}-4x+1=x^2-4x+1\end{bmatrix}[/tex]
[tex]x^2=0, x = 0[/tex]
[tex]y=-4\cdot \:0+1 = 1[/tex]
[tex]y=1,\:x=0[/tex]
The point of intersection would be (0, 1)
y = -4x + 3 [intersects parabola at 2 points]
Check:
[tex]\begin{bmatrix}y=\left(x-2\right)^2-3\\ y=-4x+3\end{bmatrix}[/tex]
Subtract the 2 equations,
[tex]y=\left(x-2\right)^2-3\\-\\\underline{y=-4x+3}\\y-y=\left(x-2\right)^2-3-\left(-4x+3\right),\\0=x^2-2\\x = \sqrt{2}, x = - \sqrt{2} \\\\\mathrm{Plug\:the\:solutions\:}x=\sqrt{2},\:x=-\sqrt{2}\mathrm{\:into\:}y=\left(x-2\right)^2-3[/tex]
[tex]\begin{pmatrix}x=\sqrt{2},\:&y=3-4\sqrt{2}\\ x=-\sqrt{2},\:&y=3+4\sqrt{2}\end{pmatrix}[/tex]
Therefore the points of intersection are (√2, 3-4√2) and (-√2, 3+4√2) respectively
And finally we have the equation y = - 4x. It doesn't intersect the parabola.
c) The y-intercepts of such graphs are all less than or equal to 0.
