Question

Find the x- and y- intercepts of parabola y=5x^2-16x+10

Answer 1

Y-INTERCEPT

`y = 5x^2 - 16x + 10`

The y-intercept is where the equation/curve/parabola cosses the y-axis.

The y-axis is where x = 0. (The x-axis is where y = 0)

To find the y-intercept:

`\text{y-axis} \rightarrow \text{x = 0} \rightarrow y = 5(0)^2 -16(0) + 10 = 10`

The y-intercept must be at (0, 10)

X-INTERCEPT (ROOTS/SOLUTIONS)

`y = 5x^2 - 16x + 10\\\text{make it equal 0}\\y = 0\\\therefore 5x^2 - 16x + 10 = 0`

We need to use the quadratic formula

The quadratic formula helps us find what values of
`x` make the equation = 0

Quadratic formula:
`x=(-b\pm√(b^2-4ac))/(2a)`

`x=(-(-16) + √((-16)^2-4(5)(10)))/(2(5))\\\\x = (16 + √(256-200))/(10)\\x = (16 + √(56))/(10)\\x = (16 + 2√(14))/(10)\\x = (8 + √(14))/(5)\\\\\\x=(-(-16) - √((-16)^2-4(5)(10)))/(2(5))\\\text{doing the same thing...}\\\text{end up with...}\\x = (8 - √(14))/(5)\\`

The x-intercepts are at:

`((8 + √(14))/(5), 0)\\((8 - √(14))/(5), 0)`