Question

Find the vertex of the parabola y = -5x² - 6x + 15/2

Answer 1

`\\ \sf\longmapsto y=-5x^2-6x+15/2`

`\\ \sf\longmapsto -5x^2-6x-y+15/2=0`

• Multiply 2 with each term

`\\ \sf\longmapsto -10x^2-12x-2y+15=0`

• Convert to vertex form

`\\ \sf\longmapsto y=-5(x^2+(6)/(5)x+(9)/(25))+(93)/(10)`

`\\ \sf\longmapsto y=-5(x^2+2(3)/(5)x+((3)/(5))^2)+(93)/(10)`

`\\ \sf\longmapsto y=-5(x+(3)/(5))^2+(93)/(10)`

Compare to vertex form of parabola

`\boxed{\sf y=a(x-h)^2+k}`

Now

• h=-3/5
• k=93/10

Answer 2

• (- 0.6, 9.3)

Explanation:

Given parabola:

• y = -5x² - 6x + 15/2

The vertex is determined by x = - b/2a:

• x = - (-6)/(2*(-5)) = -0.6

Find y- coordinate of vertex:

• y = - 5(0.6)² - 6(- 0.6) + 15/2 = 9.3