Answer to Question #286242 in Calculus for Ima

Question

01) Identify the X intercepts of this quadratic function y = 2x- x²

(02) Find the coordinates of maximum point of y = -4x + 8x – 2 by using equation y by using equetion method.

(03) Graph the quadratic function y = -3x +x+1 and you are required to answer the followings by using the graph:

(i) Coordinates of maximum point

(ii) Axis of symmetry

Answer 1

(01)

X-intercept(s)

$$y=0=> 2x- x^2=0$$

$$x(2-x)=0$$

$$x_1=0, x_2=2$$

X-intercept(s): $(0, 0), (2, 0).$

(02)

$$y=-3x^2+8x-2$$

$$y=-3(x^2-2(\dfrac{4}{3})x+(\dfrac{4}{3})^2)+\dfrac{16}{3}-2$$

$$y=-3(x-\dfrac{4}{3})^2+\dfrac{10}{3}$$

The maximum point is $(\dfrac{4}{3}, \dfrac{10}{3}).$

(03)

$$y = -3x^2 +x+1$$ $$y=-3(x^2-2(\dfrac{1}{6})x+(\dfrac{1}{6})^2)+\dfrac{1}{12}+1$$ $$y=-3(x-\dfrac{1}{6})^2+\dfrac{13}{12}$$

The maximum point is $(\dfrac{1}{6}, \dfrac{13}{12}).$

Axis of symmetry: $x=\dfrac{1}{6}$

y-intercept

$$x=0=> y=-3(0)^2+0+1=1$$

y-intercept: $(0, 1).$

X-intercept(s)

$$y=0=> -3x^2+x+1=0$$ $$D=(1)^2-4(-3)(1)=13$$ $$x_{1,2}=\dfrac{-1\pm \sqrt{13}}{2(-3)}$$

$$x_{1}=\dfrac{1- \sqrt{13}}{6}, x_{2}=\dfrac{1+\sqrt{13}}{6}$$

X-intercept(s): $(\dfrac{1- \sqrt{13}}{6}, 0), (\dfrac{1+\sqrt{13}}{6}, 0).$

(i) The maximum point is $(\dfrac{4}{3}, \dfrac{10}{3}).$

(ii) Axis of symmetry: $x=\dfrac{1}{6}$