Question

Find the x-intercepts of the parabola with
vertex (1,1) and y-intercept (0,-3).
Write your answer in this form: (X1,y,),(X2,).
If necessary, round to the nearest hundredth.​

Answer 1

Answer:

$(\frac{1}{2},0),(\frac{3}{2},0)$

Step-by-step explanation:

The vertex form of a parabola is given by:

$y=a(x-h)^2+k$, where V(h,k) is the vertex of the parabola.

The given parabola has vertex (1,1).

This implies that: $h=1,k=1$.

Put these values into the vertex form equation.

$\implies y=a(x-1)^2+1$

The y-intercept of this parabola is: (0,-3).

This point lies on the parabola hence it must satisfy its equation.

$\implies -3=a(0-1)^2+1$

$\implies -3=a(-1)^2+1$

$\implies -3=a(1)+1$

$\implies -3=a+1$

$\implies -3-1=a$

$\implies -4=a$

The equation now becomes

$\implies y=-4(x-1)^2+1$

To find the x-intercept, put y=0 into the equation:

$\implies -4(x-1)^2+1=0$

$\implies -4(x-1)^2=-1$

Divide through by -4.

$\implies \frac{-4(x-1)^2}{-4}=\frac{-1}{-4}$

$\implies (x-1)^2=\frac{-1}{-4}$

$\implies (x-1)^2=\frac{1}{4}$

Take plus or minus square root of both sides.

$\implies x-1=\pm \sqrt{\frac{1}{4}}$

$\implies x-1=\pm \frac{1}{2}$

$\implies x=1\pm \frac{1}{2}$

$\implies x=1-\frac{1}{2}$ or $\implies x=1+ \frac{1}{2}$

$\implies x=\frac{1}{2}$ or $\implies x=1 \frac{1}{2}$

Therefore the x-intercepts are:

$(\frac{1}{2},0),(\frac{3}{2},0)$

To the nearest hundredth, we have $0.50,0),(1.50,0)$