- Find (i) the vertex and (ii) the equation of the line of symetry of each of the following quadratic graphs.
-
y = 1(x -2)² + 3
according to the formula, the quadratic equation is in the Vertex form:
(i) its vertex is v(h, k), with h = -(-2) and k = 3 ==> v(2, 3)
1 > 0, parabola opens upwards => Minimum
(ii) its line of symetry is x = 2 (see grah)
- y = 1(x -3)² + c
(i) Vertex v(3, c)
with c undefined constant, it acts like a parameter!!!
1 > 0, parabola opens upwards => Minimum
(ii) The line of symetry of the equation is x = c, c can take any value! - y = 1(x -p)² + q
(i) h = p and k = q ==> vertex v(p, q)
p, q constant values;
1 > 0, parabola opens upwards => Minimum
(ii) Line of symetry x = p, p constant. - y = (ax + b)² + c
first bring the equation to a vertex form (put a as factor as in the vertex form): ==> y = a(x + b/a) + c (vertex form)
(i) h = −b/a, k = c, and the vertex: v(−b/a, c), a, b and c constant.
a > 0, parabola opens upwards => Minimum
(ii) Line of symetry x = −b/a, a and b constant. - Find (i) the least (or, if appropriate, the greatest) value of each of the following quadratic expressions and (ii) the value of x for which this occurs.
- Vertex form of a quadratic function: f(x) = a(x - h)² + k.
- a determines how the parabola opens:
- a > 0, it opens upwards;
- a < 0, it opens downwards. - h and k are the values of the vertex.
- The greatest or least value of a quadratic function equals k.
- The greatest or least value of a quadratic function occurs at x = h.
- Vertex: v(h, k)
- y = 1(x + 2)² - 1
(x + 2)² - 1 is smallest for x + 2 = 0
x + 2 = 0 when x = −2
==> the function will have its smallest output when x = −2
so let's find the output now:
f(-2) = (-2 + 2)² - 1 = (0)² −1 = −1
So the least value of y = 1(x + 2)² - 1 is k = −1 and it occurs at x = −2, and the parabola opens upwards for a = 1 > 0. - y = 1(x - 1)² + 2 a = 1 > 0, the parabola opens upwards
- y = 5 - 1(x + 3)² or y = −1(x + 3)² + 5
- y = 1(2x + 1)² − 7 by factorizing 2: 2(x + 1/2)² − 7
- y = 3 − 2(x − 4)² or y = − 2(x − 4)² + 3 a = −2 < 0, the parabola opens downwards
- y = 1(x + p)² + q, with p and q constant. a = 1 > 0, the parabola opens upwards
- y = 1(x − p)² − q, with p and q constant. a = 1 > 0, the parabola opens upwards
- y = r −1(x − t)² or y = −1(x − t)² + r, with t and r constant. a = −1 < 0, the parabola opens downwards
- y = c −1(ax + b)² or y = −1(ax + b)² + c, with a, b and c constant. factorize by a: −a(x + b/a) + c (vertex form)
- Solve the following quadratic equations. Leave surds in your answer.
- (x − 3)² − 3 = 0 | +3 (on both side) (x − 3)² = 3 | √ (square root it)
- (x + 2)² − 4 = 0 | +4 (x + 2)² = 4 | √ (square root it)
- 2(x + 3)² = 5 | ÷2 (x + 3)² = 5/2 | √
- (3x − 7)² = 8 | √ 3x − 7 = ± √8 | +7
- (x + p)² − q = 0 | +q (x + p)² = q | √
- a(x + b)² − c = 0 | +c a(x + b)² = c | ÷a
- Express the following in completed square form
- x² + 2x +2 (by completing the square) (x² + 2x + 1) + 2 − 1
- x² −8x − 3 (completing the square) (x² − 8x + 16)² − 3 − 16
- x² + 3x − 7 (completing the square) (x² + 3x + 2.25)² − 7 − 2.25
- 5 − 6x + x², better x² − 6x + 5 (x² − 6x + 9) + 5 − 9
- x² + 14x + 49 (This is a completed square Form) ==> (x + 7)²
- 2x² + 12x − 5 | ÷2 (divide by 2 to bring the equation to a regular quadratic form) x² + 6x − 5/2 (complete the square)
- 3x² − 12x + 3 | ÷3 x² − 4x + 1 (complete the square)
- 7 − 8x − 4x² | ÷4 <==> −x − 2x + 7/4 | ×(−1)
- 2x² + 5x − 3 | ÷2 x² + 5/2 − 3/2 | (complete the square)
- Use the completed square form to factorize the following expressions
- x² − 2x − 35 | (complete the square)
- x² − 14x − 176 | (complete the square)
- x² + 6x − 432 | (complete the square)
- 6x − 5x − 6 | ÷6
- 14 + 45x − 14x² <==> − 14x² + 45x + 14 | − ÷14 x² − 45x/14 − 1 | (complete the square)
- 12x² + x − 6 | ÷12 x² − x/12−1/2 | (complete the square)
The same method can be applied to b., c., d., e. and f.
| To the top |
| Core Lesson |
==> its least value is k = 2 and it occurs at x = 1.
a = −1 < 0, the parabola opens downwards
==> its greatest value is k = 5 and it occurs at x = −3.
a = 2 > 0, the parabola opens upwards
==> its least value is k = −7 and it occurs at x = −1/2.
==> the least value of the function is k = 3 and it occurs at x = 4.
==> the greatest value of the function is k = q and it occurs at x = −p.
==> the greatest value of the function is k = −q and it occurs at x = p.
==> the least value of the function is k = r and it occurs at x = t.
−a < 0, the parabola opens downwards
==> the least value of the function is k = c and it occurs at x = −b/a.
x − 3 = ± √3 | +3
==> x = 3 ± √3
x + 2 = ± 2
==> x = − 2 ± √2
x + 3 = ± √5/2 | −3
==> x = − 3 ± √5/2
3x = 7 ± 2√2 | ÷3 ==> x = 7/3 ± 2√2/3
x + p = ± √q | −p
==> x = − p ± √q
(x + b)² = c/a | √
x + b = ± √c/a | −b
==> x = − b ± √c/a.
| To the top |
==> (x + 1)² + 1
==> (x − 4)² − 19
==> (x + 1.5)² − 9.25
==> (x − 3)² − 4 or (x − 3)² − 2²
(x² + 6x + 9) − 5/2 − 9
==> (x + 3)² − 23/2
(x² − 4x + 4) + 1 − 4
==> (x − 2)² − 3
<==> x + 2x −7/4 | (complete the square)
(x² + 2x + 1) −7/4 − 1
==> (x + 1) −11/4
(x² + 5/2 + 1.56) − 3/2 − 1.56
==> (x + 1.25) − 3.06
(x − 1)² − 36 <==> (x − 1)² − 6² | √ (square it)
±(x + 1) ± 6
==> (x + 1 − 6)(x + 1 + 6)
==> (x − 5)(x + 7)
==> (x − 7 + 15)(x − 7 − 15)
==> (x + 8)(x − 22)
(x + 3)² − 21² | √
==> (x + 3 − 21)(x + 3 + 21)
==> (x − 18)(x + 24)
(x² − 5x/6 + 0.18) − 1 − 0.18 (x − 0.42)² − 1.18 <==> (x − 0.42)² − ±1.08 | √
==> (x − 0.42 − 1.09)(x − 0.42 + 1.09)
==> (x − 1.5)(x + 0.67)
(x² − 3.2x + 1.6²) − 1 − 1.6²
(x − 1.6)² − 3.5 <==> (x − 1.6)² − 1.9² | √
==> (x − 1.6 − 1.9)(x − 1.6 + 1.9)
==> (x − 3.5)(x − 0.3)
(x² − x/12 + 0.041²) − 0.5 − 0.041²
(x − 0.041)² − 0.5 <==> (x − 0.041)² ± 0.7²
==> (x − 0.041 − 0.7)(x − 0.041 + 0.7)
==> (x − 0.74)(x + 0.66)
| To the top |
| Back to Exercises |
| Back to Lesson |
