Using a directrix of y = −2 and a focus of (2, 6), what quadratic function is created? f(x)=−(x−2)2−2 f(x)=(x−2)2+2 f(x)=(x−2)2−2 of f(x)=−(x+2)2−2

Question

Using a directrix of y = −2 and a focus of (2, 6), what quadratic function is created?  f(x)=−(x−2)2−2  f(x)=(x−2)2+2  f(x)=(x−2)2−2  of f(x)=−(x+2)2−2

Florence Pittman · Accepted Answer

To find equation of quadratic function Given, directrix, y= -2 And focus= (2, 6) We know, from any point (x, y) on the parabola the focus and directrix are equidistant 1. Distance of(x, y) from (2, 6)= distance of directrix, y= -2 from (x, y) ((x−2)2+(y−6)2)0.5=|y−(−2)| {using distance formula, if (x1,y1) and (x2,y2) are two points then dist between them is ((x1−x2)2+(y1−y2)2)0.5} ((x−2)2+(y−6)2)0.5=|y+2| Squaring both sides, we get (((x−2)2+(y−6)2)0.5)2=(|y+2|)2 (x−2)2+(y−6)2=(y+2)2 (y−6)2−(y+2)2=−(x−2)2 Y2+36–12y–(y2+4+4y)=−(x−2)2 Y2+36–12y−y2−4–4y=−(x−2)2 −16y+32=−(x−2)2 (x−2)2+32=16y 16y=(x−2)2+32 Y=116(x−2)2+3216 Y=116(x−2)2+2 Hence, the quadratic function is F(x)=116(x−2)2+2

Using a directrix of y = −2 and a focus

Answered question

Answer & Explanation

New Questions in Algebra I