Die hoekpunt van 'n kwadratiese vergelyking of parabool is die hoogste of laagste punt van die vergelyking. Dit lê ook op die simmetrievlak van die hele parabel; alles wat links van die parabool lê, is 'n volledige spieëlbeeld van alles wat regs is. As u die hoekpunt van 'n kwadratiese vergelyking wil vind, kan u die hoekpuntformule gebruik of die vierkant voltooi.
Stappe
Metode 1 van 2: Die gebruik van die draaikolkformule
Stap 1. Identifiseer die waardes van a, b en c
In 'n kwadratiese vergelyking is die x2 { displaystyle x^{2}}
term = a, the x{displaystyle x}
term = b, and the constant term (the term without a variable) = c. Let's say you're working with the following equation: 'y=x2+9x+18{displaystyle y=x^{2}+9x+18}
. In this example, a{displaystyle a}
=
Step 1., b{displaystyle b}
=
Step 9., and c{displaystyle c}
=
Step 18..
Stap 2. Gebruik die hoekpuntformule om die x-waarde van die hoekpunt te vind
Die hoekpunt is ook die simmetrie -as van die vergelyking. Die formule vir die vind van die x-waarde van die hoekpunt van 'n kwadratiese vergelyking is x = −b2a { displaystyle x = { frac {-b} {2a}}}
. Plug in the relevant values to find x. Substitute the values for a and b. Show your work:
- x=−b2a{displaystyle x={frac {-b}{2a}}}
- x=−(9)(2)(1){displaystyle x={frac {-(9)}{(2)(1)}}}
- x=−92{displaystyle x={frac {-9}{2}}}
Stap 3. Koppel die x { displaystyle x}
value into the original equation to get the y{displaystyle y}
value.
Now that you know the x{displaystyle x}
value, just plug it in to the original formula for the y{displaystyle y}
value. You can think of the formula for finding the vertex of a quadratic function as being (x, y)=[(−b2a), f(−b2a)]{displaystyle (x, y)=\left[({frac {-b}{2a}}), f({frac {-b}{2a}})\right]}
. This just means that to get the y{displaystyle y}
value, you have to find the x{displaystyle x}
value based on the formula and then plug it back into the equation. Here's how you do it:
- y=x2+9x+18{displaystyle y=x^{2}+9x+18}
- y=(−9)(2)2+9(−9)(2)+18{displaystyle y={frac {(-9)}{(2)}}^{2}+9{frac {(-9)}{(2)}}+18}
- y=814−812+18{displaystyle y={frac {81}{4}}-{frac {81}{2}}+18}
- y=814−1624+724{displaystyle y={frac {81}{4}}-{frac {162}{4}}+{frac {72}{4}}}
- y=(81−162+72)4{displaystyle y={frac {(81-162+72)}{4}}}
-
y=−94{displaystyle y={frac {-9}{4}}}
Stap 4. Skryf die x { displaystyle x} neer
and y{displaystyle y}
values as an ordered pair.
Now that you know that x=−92{displaystyle x={frac {-9}{2}}}
and y=−94{displaystyle y={frac {-9}{4}}}
just write them down as an ordered pair: (−92, −94){displaystyle ({frac {-9}{2}}, {frac {-9}{4}})}
. The vertex of this quadratic equation is (−92, −94){displaystyle ({frac {-9}{2}}, {frac {-9}{4}})}
. If you were to draw this parabola on a graph, this point would be the minimum of the parabola, because the x2{displaystyle x^{2}}
term is positive.
Method 2 of 2: Completing the Square
Stap 1. Skryf die vergelyking neer
Die voltooiing van die vierkant is nog 'n manier om die hoekpunt van 'n kwadratiese vergelyking te vind. Vir hierdie metode, as u aan die einde kom, kan u u x- en y -koördinate dadelik vind, in plaas daarvan om die x -koördinaat weer aan te sluit by die oorspronklike vergelyking. Gestel jy werk met die volgende kwadratiese vergelyking: x2+4x+1 = 0 { displaystyle x^{2}+4x+1 = 0}
Stap 2. Verdeel elke term deur die koëffisiënt van die x2 { displaystyle x^{2}}
term.
In this case, the coefficient of the x2{displaystyle x^{2}}
term is 1, so you can skip this step. Dividing each term by
Step 1. would not change anything. Dividing each term by 0, however, will change everything.
Stap 3. Beweeg die konstante term na die regterkant van die vergelyking
Die konstante term is die term sonder 'n koëffisiënt. In hierdie geval is dit so
Stap 1.. Mov
Stap 1. na die ander kant van die vergelyking deur subtractien
Stap 1. van beide kante. Hier is hoe jy dit doen:
- x2+4x+1 = 0 { displaystyle x^{2}+4x+1 = 0}
- x2+4x+1−1=0−1{displaystyle x^{2}+4x+1-1=0-1}
- x2+4x=−1{displaystyle x^{2}+4x=-1}
Stap 4. Voltooi die vierkant aan die linkerkant van die vergelyking
Om dit te doen, vind (b2) 2 { displaystyle ({ frac {b} {2}})^{2}}
and add the result to both sides of the equation. Plug in
Step 4. for b{displaystyle b}
since4x{displaystyle 4x}
is the b-term of this equation.
-
(42)2=22=4{displaystyle ({frac {4}{2}})^{2}=2^{2}=4}
. Now, add
Step 4. to both sides of the equation to get the following:
- x2+4x+4=−1+4{displaystyle x^{2}+4x+4=-1+4}
- x2+4x+4=3{displaystyle x^{2}+4x+4=3}
Stap 5. Faktor die linkerkant van die vergelyking
Nou sal jy sien dat x2+4x+4 { displaystyle x^{2}+4x+4}
is a perfect square. It can be rewritten as (x+2)2=3{displaystyle (x+2)^{2}=3}
Stap 6. Gebruik hierdie formaat om die x { displaystyle x} te vind
and y{displaystyle y}
coordinates.
You can find your x{displaystyle x}
coordinate by simply setting (x+2)2{displaystyle (x+2)^{2}}
equal to zero. So when (x+2)2=0{displaystyle (x+2)^{2}=0}
what would x{displaystyle x}
have to be? The variable x{displaystyle x}
would have to be -
Step 2. to balance out the +2, so your x{displaystyle x}
coordinate is - 2. Your y-coordinate is simply the constant term on the other side of the equation. So, y=3{displaystyle y=3}
. You can also do a shortcut and just take the opposite sign of the number in parentheses to get the x-coordinate. So the vertex of the equation x2+4x+1=(−2, −3){displaystyle x^{2}+4x+1=(-2, -3)}
Video - Deur hierdie diens te gebruik, kan sommige inligting met YouTube gedeel word.
<iframe
Wenke
- Identifiseer korrek a, b en c.
- Wys altyd u werk. Dit help nie net diegene wat u merk, dat u weet wat u doen nie, maar dit help u ook om te sien waar u foute begaan.
- Die volgorde van bedrywighede moet gevolg word vir 'n korrekte resultaat.
Waarskuwings
- Wys en kyk na u werk!
- Maak seker dat u weet wat a, b en c is - as u dit nie doen nie, is die antwoord verkeerd.
- Moenie jouself stres nie - dit kan oefening verg.
