Question

given a quadratic equation h(x)=3x²-12x+7

a) find the vertex algebraically

b) write the equation in vertex form

c) find the x and y-intercepts algebraically

Answer 1

a). The given quadratic equation is h(x)=3x²-12x+7 = 3(x²-4x)+7 = 3(x²-4x +4)+7-3*4 =3(x-2)² -5.

This is the vertex form of the equation of a parabola. The vertex is at the point (2,-5).

b).The vertex form of the equation is h(x) = 3(x-2)² -5.

c). The x-intercept is where y = 0. On substituting y = h(x) = 0, we get 3(x-2)² -5 = 0 or, 3(x-2)² =5 or, (x-2)² = 5/3 so that x-2 = ±?(5/3). Hence x = 2±?(5/3).Thus, the x-intercepts are 2-?(5/3) and 2+?(5/3).

The y-intercept is where x= 0. On substituting x = 0, we get y = h(x) = 3*0²-12*0+7 = 7. Thus, the y-intercept is 7.