Respuesta :
1. Consider the functions f(x)=-4x+3 and g(x)=[tex]- x^{2} +4x-3[/tex]
the graph of f is a line (f is a linear polynomial function) so we can use 2 points to draw it, say (0, 3) and (2, -5)
the graph of g is a parabola (g is a quadratic polynomial function), so we can factorize the expression as follows:
[tex]- x^{2} +4x-3=-(x^{2} -4x+3)=-(x-3)(x-1)[/tex]
so the roots of g are 1 and 3, the axis of symmetry is the vertical line through (2,0), so the vertex is calculated at x=2, -(2-3)(2-1)=-(-1)(1)=1. Vertex is the point (2, 1)
2. check the picture attached. Clearly there are 2 solutions, at the 2 intersections of the graphs.
3. Algebraically, the solutions are at :
[tex]-4x+3 =- x^{2} +4x-3[/tex]
[tex]0=- x^{2} +8x-6[/tex]
[tex] x^{2} -8x+6=0[/tex]
[tex]x^{2} -8x+6=x^{2} -2*4x+16-16+6=(x-4)^{2}-10=0 [/tex]
[tex](x-4)^{2}=10[/tex]
[tex]x-4= +\sqrt{10} [/tex] or [tex]x-4= -\sqrt{10} [/tex]
[tex]x=4 +\sqrt{10} [/tex] or [tex]x=4 -\sqrt{10} [/tex]
the graph of f is a line (f is a linear polynomial function) so we can use 2 points to draw it, say (0, 3) and (2, -5)
the graph of g is a parabola (g is a quadratic polynomial function), so we can factorize the expression as follows:
[tex]- x^{2} +4x-3=-(x^{2} -4x+3)=-(x-3)(x-1)[/tex]
so the roots of g are 1 and 3, the axis of symmetry is the vertical line through (2,0), so the vertex is calculated at x=2, -(2-3)(2-1)=-(-1)(1)=1. Vertex is the point (2, 1)
2. check the picture attached. Clearly there are 2 solutions, at the 2 intersections of the graphs.
3. Algebraically, the solutions are at :
[tex]-4x+3 =- x^{2} +4x-3[/tex]
[tex]0=- x^{2} +8x-6[/tex]
[tex] x^{2} -8x+6=0[/tex]
[tex]x^{2} -8x+6=x^{2} -2*4x+16-16+6=(x-4)^{2}-10=0 [/tex]
[tex](x-4)^{2}=10[/tex]
[tex]x-4= +\sqrt{10} [/tex] or [tex]x-4= -\sqrt{10} [/tex]
[tex]x=4 +\sqrt{10} [/tex] or [tex]x=4 -\sqrt{10} [/tex]