Question

Consider the quadratic function, f(x) = 3x^2-12x + 7a) Which direction does the parabola open?b) What is the equation for the axis of symmetry? c) What are the coordinates of the vertex?

Answer 1

Answer:

Given quadratic function,

$f\mleft(x\mright)=3x^2-12x+7$

a) To find direction does the parabola open.

First we simplify the given equation in the standard equation of parabola,

$f\mleft(x\mright)=3\lparen x^2-4x)+7$$f\mleft(x\mright)=3\operatorname{\lparen}x^2-4x+4-4)+7$$f\mleft(x\mright)=3\operatorname{\lparen}x^2-4x+2^2)-12+7$$f\mleft(x\mright)=3\left(x-2\right)^2-5$

we know that, y=f(x), using this we get,

$y+5=3(x-2)^2$$\left(x-2\right)^2=\frac{1}{3}\left(y+5\right)$

From the standard form of the equation we have that,

$(x-h)^2=4a(y-k)$

(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,

$\left(x-2\right)^2=\frac{1}{3}\left(y+5\right)$

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,

$x=2$

c) To find the coordinates of the vertex,

The equation of the parabola is,

$\left(x-2\right)^2=\frac{1}{3}\left(y+5\right)$

Vertex is (2,-5).