Question

PLEASE HELP ME ILL GIVE 80 POINTS using only algebra (no graphing) please find the x-intercepts (if they exist), y-intercept, vertex, and axis of symmetry for this quadratic: Y= -4x^2 + 24x - 10

Answer 1

find the x intercepts:

-4x²+24x-10=0

2x²-12x+5=0

formula:

`\frac{ - b + \sqrt{b { }^(2) - 4ac } }{2a}`

`\frac{ - 12 \sqrt{12 {}^(2) - 4(2)(5) } }{2(2)}`

the two results are:

X1= (6-√26)/2

X2= (6+√26)/2

find the y intercepts:

substitute x for 0:

y=-4(0)²+24(0)-10

y=-10

find the vertex:

find the coordinate x of the vertex using the formula:

`x = - (b)/(2a)`

`x = - (24)/(2 * ( - 4))`

x=3

now, substitute x for 3:

y=-4(3)²+24(3)-10

y=26

the vertex is (3,26)

find the axis of symmetry:

use the formula:

`x = - (b)/(2a)`

`x = - (24)/(2 * ( - 4))`

x=3

Answer 2

Sure! Let's start with the x-intercepts. To find the x-intercepts of a quadratic equation, we need to set y=0 and solve for x. So, we have:

-4x^2 + 24x - 10 = 0

We can simplify this equation by dividing all terms by -2:

2x^2 - 12x + 5 = 0

Now we can use the quadratic formula to solve for x:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 2, b = -12, and c = 5.

Plugging these values into the formula, we get:

x = (12 ± sqrt(144 - 40)) / 4

x = (12 ± sqrt(104)) / 4

x = (12 ± 2sqrt(26)) / 4

x = 3 ± (1/2)sqrt(26)

So the x-intercepts are approximately:

(-0.5359, 0) and (6.5359, 0)

To find the y-intercept, we need to set x=0 and solve for y:

y = -4(0)^2 + 24(0) - 10

y = -10

So the y-intercept is (0,-10).

To find the vertex of the parabola, we can use the formula:

x = -b / 2a

y = f(x)

where f(x) is the quadratic function.

In this case, a = -4, b = 24, and c = -10.

So we have:

x = -24 / (2(-4))

x = 3

y = -4(3)^2 + 24(3) - 10

y = -22

So the vertex is (3,-22).

Finally, to find the axis of symmetry, we can use the formula:

x = -b / 2a

which we already used to find the vertex. So the axis of symmetry is x=3.

I hope that helps! :)