Question

The vertex of a parabola is (-1.5, -12.5), and its y-intercept is (0, -8).
The x-intercepts of the parabola are

Answer 1

Answer:

The x-intercept are x=-4,1.                    

Step-by-step explanation:

Given : The vertex of a parabola is (-1.5, -12.5), and its y-intercept is (0, -8).

To find : The x-intercepts of the parabola are ?

Solution :

The general form of the parabola equation is  

$y=a(x-h)^2+k$ ......(1)

where, a is the initial value and (h,k) are vertex of the parabola.

We have given the vertex (h,k)=(-1.5, -12.5)

Substitute in equation (1),

$y=a(x-(-1.5))^2+(-12.5)$

$y=a(x+1.5)^2-12.5$ ......(2)

Now, The y-intercept is (0,-8)

Substitute in equation (2),

$-8=a(0+1.5)^2-12.5$

$-8+12.5=a(1.5)^2$

$4.5=2.25a$

$a=\frac{4.5}{2.25}$

$a=2$

Put back in equation (2),

$y=2(x+1.5)^2-12.5$

To find the x-intercept put y=0 in above equation,

$0=2(x+1.5)^2-12.5$

$0=2(x^2+3x+2.25)-12.5$

$0=2x^2+6x+4.5-12.5$

$0=2x^2+6x-8$

$x^2+3x-4=0$

Solve by middle term split,

$x^2+4x-x-4=0$

$x(x+4)-1(x+4)=0$

$(x+4)(x-1)=0$

$x+4=0,x-1=0$

$x=-4,1$

Therefore, The x-intercept are x=-4,1.

Answer 2

Answer:

x=-4,1

Step-by-step explanation: