# What is the highest point of y = -4x^2 + 3x - 8.

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### 3 Answers

`y = -4x^2 + 3x - 8`

`y = -4(x^2-3/4x+2)`

`y = -4(x^2-3/4x+9/64-9/64+2)`

`y = -4(x^2-3/4x+9/64)+4xx9/64-8`

`y = -4(x-3/8)^2+9/16-8`

`y = -4(x-3/8)^2-119/16`

`y = -4(x-3/8)^2-7.4375`

`(x-3/8)^2>= 0` always. So minimum of `(x-3/8)^2 = 0` .

`-4(x-3/8)^2 <= 0` . So maximum of `(x-3/8)^2` comes when `(x-3/8)^2= 0.`

`(x-3/8)^2 = 0`

`x-3/8 = 0`

`x = 3/8`

`y = -4(x-3/8)^2-7.4375`

`y = -4(0)-7.4375`

`y = -7.4375`

*So the highest point is (3/8,-7.4375)*

**Sources:**

The graph of y = -4x^2 + 3x - 8 is that of a parabola. The highest point of the graph lies at the solution x = a, of y' = 0.

y' = -8x + 3

-8x + 3 = 0

=> x = 3/8

y = -4*(3/8)^2 + 3*(3/8) - 8

= -7.4375

**The highest point of the graph is (0.375, -7.4375)**

y = -4x^2 + 3x - 8.

`y=-4x^2+3x-8`

`y=-4(x^2-(3x)/4+2)`

`y=-4(x^2-2(3/8)x+9/64+119/64)`

`y=-4(x-3/8)^2-119/16`

`y+119/16=-4(x-3/8)^2`

`` `(-1/4)(y+119/16)=(x-3/8)^2`

Graph is parabola and defined if `y+119<=0`

`y<=-119/16`

This mean graph is opening downword. So maximum exist.

maximum point (3/8,-119/16) i.e vertex of parabola.