Ex: Quadratic Function Application – Horizontal Distance and Vertical Height

Ex: Quadratic Function Application – Horizontal Distance and Vertical Height


– THE EQUATION Y=-ONE
SIXTEENTH X SQUARED + 4X +3 MODELS THE HEIGHT OF AN ARROW WHERE X IS THE HORIZONTAL
DISTANCE IN FEET FROM THE POINT
THE ARROW IS SHOT. SO Y REPRESENTS THE HEIGHT
IN FEET AND X REPRESENTS
THE HORIZONTAL DISTANCE TRAVELED IN FEET. WE WANT TO ANSWER
THE FOLLOWING THREE QUESTIONS. NUMBER ONE, HOW HIGH
IS THE ARROW WHEN IT IS SHOT, NUMBER TWO, WHAT IS THE
MAXIMUM HEIGHT OF THE ARROW, AND NUMBER THREE, HOW FAR HORIZONTALLY
DOES THE ARROW TRAVEL BEFORE HITTING THE GROUND. SO LET’S GO THROUGH THESE ONE
AT A TIME. TO ANSWER THE FIRST QUESTION OF HOW HIGH THE ARROW IS
WHEN IT IS SHOT, RIGHT AT THE INSTANT
THE ARROW IS SHOT, THE ARROW HAS NOT TRAVELED
HORIZONTALLY THEREFORE, TO ANSWER
THIS QUESTION, WE’RE GOING TO SET X EQUAL
TO 0 AND THEN SOLVE FOR Y. SO IF WE LET X=0,
WE WANT TO EVALUATE Y OF 0 WHICH WOULD BE -ONE SIXTEENTH
x 0 SQUARED + 4X 0 +3 SO Y OF 0 IS JUST GOING TO BE
0 + 0 +3 OR 3. THEREFORE THE ARROW STARTS
AT A HEIGHT OF THREE FEET. NUMBER TWO, WE WANT TO DETERMINE THE
MAXIMUM HEIGHT OF THE ARROW. NOTICE OUR EQUATION
IS A QUADRATIC EQUATION AND THEREFORE THE GRAPH
IS A PARABOLA AND BECAUSE “A” IS NEGATIVE, IT’S A PARABOLA
THAT OPENS DOWN OR LOOKS SOMETHING LIKE THIS. SO THE MAXIMUM HEIGHT OF
THE ARROW WOULD BE REPRESENTED BY THE VERTEX. SO WELL FIND THE VERTEX WHERE THE X COORDINATE
WILL REPRESENT THE HORIZONTAL DISTANCE
TRAVELED AND THE Y COORDINATE
WILL BE THE MAXIMUM HEIGHT. REMEMBER THE COORDINATES
OF THE VERTEX ARE -B OVER 2A,
F OF -B OVER 2A. SO WE’LL START BY DETERMINING
THE X COORDINATE OF THE VERTEX USING THIS FORMULA HERE. NOTICE “A”
IS EQUAL TO -ONE SIXTEENTH,   B IS EQUAL TO POSITIVE 4,
AND WE DON’T NEED C, BUT C IS EQUAL TO 3. SO THE X COORDINATE IS GOING
TO BE EQUAL TO -B OR -4 DIVIDED BY 2 TIMES “A”
OR 2 TIMES -ONE SIXTEENTH. SO LET’S PUT THIS
TWO OVER ONE, WE’RE GOING TO HAVE -4 DIVIDED
BY THE 2 AND THE 16 SIMPLIFY, THIS SIMPLIFIES THE 1,
THIS SIMPLIFIES THE 8. SO WE HAVE -ONE EIGHTH. SO WE’LL HAVE -4 x
THE RECIPROCAL OF -ONE EIGHTH WHICH IS JUST -8,
OR -8 OVER 1. SO X IS EQUAL TO 32
OR 32 FEET. AGAIN THIS IS THE HORIZONTAL
DISTANCE TRAVELED AT THE MAXIMUM HEIGHT. SO TO FIND THE MAXIMUM HEIGHT, WE’LL NOW HAVE TO SUBSTITUTE
X=32 BACK INTO THE ORIGINAL
EQUATION. SO WE WANT TO FIND Y OF 32, WHICH IS GOING TO BE EQUAL TO
-ONE SIXTEENTH x 32 SQUARED, + 4 x 32 +3. LET’S EVALUATE THIS
ON THE CALCULATOR. SO WE HAVE -ONE SIXTEENTH,
x 32 SQUARED, + 4 x 32 +3.   SO THE MAXIMUM HEIGHT
IS EQUAL TO 67 FEET.   AND THEN
FOR THE THIRD QUESTION, WE WANT TO DETERMINE
HOW FAR HORIZONTALLY DOES THE ARROW TRAVEL
BEFORE HITTING THE GROUND. WELL WHEN THE ARROW
HITS THE GROUND, THE VERTICAL HEIGHT
WOULD BE 0, WHICH MEANS Y WOULD BE
EQUAL TO 0. SO WE’RE GOING TO SET Y
EQUAL TO 0 SO WE’D HAVE THE EQUATION 0=-ONE SIXTEENTH X SQUARED
+ 4X +3. NOW THIS EQUATION
IS NOT GOING TO FACTOR SO WE’LL HAVE TO SOLVE THIS
USING THE QUADRATIC FORMULA. AND IT’S GOING TO BE EASIER
IF WE ELIMINATE THIS FRACTION BY MULTIPLYING BOTH SIDES
BY 16 OR IF WE WANT “A”
TO BE POSITIVE, LET’S MULTIPLY BOTH SIDES
BY A -16. REMEMBER MULTIPLYING
BOTH SIDES OF AN EQUATION BY A CONSTANT IS NOT GOING
TO CHANGE THE SOLUTIONS. SO WE’LL HAVE THE 0=THIS WOULD BE
POSITIVE 1X SQUARED – 64X AND THIS WOULD BE MINUS 48. SO NOW WE’LL USE
THE QUADRATIC FORMULA WHERE “A” IS EQUAL TO 1,
B IS EQUAL TO -64 AND C IS EQUAL TO -48. SO WE’LL HAVE X=-B OR -64,
THAT WILL BE POSITIVE 64 + OR – THE SQUARE ROOT
OF B SQUARED, WHICH IS – 64 SQUARED -4 x
“A” WHICH IS 1 x C WHICH IS – 48. HOWEVER 2 x A OR 2 x 1
WHICH IS 2. NOW LET’S GO AHEAD
AND BEGIN TO SIMPLIFY THIS. WE’LL HAVE X=64 + OR – WE’LL
COME BACK TO THE DISCRIMINATE ALL THIS IS DIVIDED BY 2. LET’S EVALUATE THIS
ON THE CALCULATOR. SO THE DISCRIMINATE IS – 64
SQUARED – 4 x 1 WHICH WE DON’T NEED BUT I’LL PUT IT IN THERE
x – 48. SO WE HAVE 4,288
UNDERNEATH THE SQUARE ROOT AND BECAUSE WE KNOW
X HAS TO BE POSITIVE, IT’S A HORIZONTAL DISTANCE, WE ONLY HAVE TO BE CONCERNED ABOUT THE POSITIVE SOLUTION
HERE WHICH WOULD BE 64 +
THE SQUARE ROOT OF 4,288 ALL DIVIDED BY 2. SO LET’S GO BACK
TO THE CALCULATOR. OUR NUMERATOR IS GOING TO BE
64 + THE SQUARE ROOT OF 4,288. NOTICE HOW WE HAVE
TWO CLOSED PARENTHESIS HERE, ONE FOR THE SQUARE ROOT
AND ONE FOR THE NUMERATOR, DIVIDED BY TWO. SO THE ARROW TRAVELS APPROXIMATELY 64.7 FEET
HORIZONTALLY BEFORE HITTING THE GROUND.   OKAY, HOPE THIS EXAMPLE HELPS.  

Leave a Reply

Your email address will not be published. Required fields are marked *