Question

How do you use the important points to sketch the graph of `y = 6x^2 + 7x - 20`?

Answer 1

Vertex: `(-0.58,-22.04)`, `y` intercept : `(0,-20)`
`x` intercepts: `(-2.5,0) and (1.33,0)`, additional point:
`(-2,-10)`

Explanation:

`y= 6 x^2+7 x -20 ; [ax^2+bx+c]:. a =6 , b=7 , c=-20`

`x` co ordinate of vertex is `x= -b/(2a)=-7/12~~ -0.583`

`y` co ordinate of vertex is `y = 6 (-7/12)^2+7(-7/12)-20` or

`y ~~ -22.042`. This is equation of a parabola of which vertex is

at `(-0.583,-22.042)`. since `a` is positive, the parabola opens

upward. y intercept : `x=0:. y =-20 or (0,-20)`

x intercept : `y=0:. 6 x^2+7 x -20` Discriminant `D= b^2-4ac`

or `D=49+480 =529`, discriminant positive, we get two real

solutions, Quadratic formula: `x= (-b+-sqrtD)/(2a)`or

`x= (-7+-sqrt 529)/12 or x = (-7+- 23)/12` or

`x = -30/12= - 2.5 , x =16/12 or x= 4/3 ~~ 1.33`

`x` intercepts are at `(-2.5,0) and (1.33,0)`

Additional point:

`x=-2 , y= 6(-2)^2+7(-2)-20= -10 or (-2,-10)`

graph{6x^2+7x -20 [-80, 80, -40, 40]} [Ans]

Answer 2

Please see the explanation below

Explanation:

The function is

`y=6x^2+7x-20`

As the coefficient of `x^2` is

`=6>0`

The curve is convex, that is, the parabola is turned upwards with a minimum.

When `x=0`, `=>`, `y=0+0-20`

This point `(0,-20)` is the intercept with the x-axis

Let's factorise

`y=(3x-4)(2x+5)`

The points of intersection with the x-axis is when `y=0`

`(3x-4)(2x+5)=0`

`=>` `{(x=4/3),(x=-5/2):}`

The points are

`(4/3,0)` and `(-5/2,0)`

Let `f(x)=6x^2+7x-20`

To determine the axis of symmetry `x=a`, apply

`f(a+h)=f(a-h)` where `h in RR`

Therefore,

`6(a+h)^2+7(a+h)-20=6(a-h)^2+7(a-h)-20`

So,

`6a^2+12ah+6h^2+7a+7h=6a^2-12ah+6h^2+7a-7h`

`24ah=-14h`

As `a!=0`

`a=14/24=-7/12`

The axis of symmetry is `x=-7/12=-0.583`

And when `x=-0.583`, `=>`, `y=-22.04`

The minimum point is `(-0.583, -22.04)`

Now, trace the curve

graph{6x^2+7x-20 [-25, 39.96, -22.7, 9.78]}