Question

Find the x-intercepts of the parabola with vertex (1,20) and y-intercept (0,16), Write your answer in this form: (X1,Y1), (X2,Y2). If necessary, round to the nearest hundredth.

Answer 1

`(-1.24,0),(3.24,0)`

Step-by-step explanation

We want to find the x intercepts of the parabola with vertex (1, 20) and y intercept (0, 16).

To do this, first find the equation of the parabola.

The vertex form of the equation of a parabola is:

`\begin{gathered} y=a(x-h)^2+k \\ \text{where (h, k) = vertex} \end{gathered}`

Therefore, we have:

`y=a(x-1)^2+20`

To find the value of a, substitute the values of x and y of the y intercept into the equation:

`\begin{gathered} 16=a(0-1)^2+20 \\ 16=a(-1)^2+20 \\ 16=a+20 \\ \Rightarrow a=16-20 \\ a=-4 \end{gathered}`

Therefore, the equation of the parabola in vertex form is:

`y=-4(x-1)^2+20`

Now, write it in standard form:

`\begin{gathered} y=-4(x^2-2x+1)+20 \\ y=-4x^2+8x-4+20 \\ y=-4x^2+8x+16 \end{gathered}`

To find the x intercepts, find the values of x when y = 0:

`\begin{gathered} -4x^2+8x+16=0 \\ \text{Solve using Quadratic formula:} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=-4;b=8;c=16 \\ x=\frac{-8\pm\sqrt[]{8^2-4(-4)(16)}}{2(-4)}=\frac{-8\pm\sqrt[]{64+256}}{-8} \\ x=\frac{-8+\sqrt[]{320}}{-8};x=\frac{-8-\sqrt[]{320}}{-8} \\ \Rightarrow x=(-8+17.89)/(-8);x=(-8-17.89)/(-8) \\ x=-1.24;x=3.24 \end{gathered}`

Therefore, the x intercepts are:

`(-1.24,0),(3.24,0)`