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Find the x-intercepts of the parabola with vertex (4,-1) and y-intercept (0,15). Write your answer in this form: (X1,Y1), (X2,72). If necessary, round to the nearest hundredth.

User Alavrik
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1 Answer

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\begin{gathered} \text{ The parabola with vertex (4,-1) and y-intercept (0,15)} \\ \\ y=a(x-4)^2+(-1) \\ y=a(x-4)^2-1 \\ \\ 15=a(-4)^2-1 \\ 15=16a-1 \\ 16=16a \\ a=1 \\ \\ \text{ The equation of the parabola is } \\ y=(x-4)^2-1 \\ \\ \text{The x-intercepts ocur when y=0} \\ \\ (x-4)^2-1=0 \\ (x-4)^2=1 \\ (x-4)=\pm\sqrt[]{1} \\ x=4\pm1 \\ \\ \text{ The two x-intercepts are:} \\ \\ (5,0)\text{ and (3,0)} \end{gathered}

User Harris
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