Respuesta :
Answer:
Given quadratic function,
[tex]f\mleft(x\mright)=3x^2-12x+7[/tex]a) To find direction does the parabola open.
First we simplify the given equation in the standard equation of parabola,
[tex]f\mleft(x\mright)=3\lparen x^2-4x)+7[/tex][tex]f\mleft(x\mright)=3\operatorname{\lparen}x^2-4x+4-4)+7[/tex][tex]f\mleft(x\mright)=3\operatorname{\lparen}x^2-4x+2^2)-12+7[/tex][tex]f\mleft(x\mright)=3\left(x-2\right)^2-5[/tex]we know that, y=f(x), using this we get,
[tex]y+5=3(x-2)^2[/tex][tex]\left(x-2\right)^2=\frac{1}{3}\left(y+5\right)[/tex]From the standard form of the equation we have that,
[tex](x-h)^2=4a(y-k)[/tex](h,k) is the vertex, and the parabola is open upward.
Hence the parabola is open upward.
b) To find the equation for the axis of symmetry.
The equation of the parabola is,
[tex]\left(x-2\right)^2=\frac{1}{3}\left(y+5\right)[/tex]Vertex is (2,-5),
To axis of symmetry passes through the vertex and parallel to y axis since it is open upward.
we get,
Axis of symmetry is,
[tex]x=2[/tex]c) To find the coordinates of the vertex,
The equation of the parabola is,
[tex]\left(x-2\right)^2=\frac{1}{3}\left(y+5\right)[/tex]Vertex is (2,-5).
