Quadratic function (solutions & vertex) minus sign trolling

[Suwandy]

I have quadratic functions: y=−(x+1)(x−7).
The goal is to find its solutions and vertex coordinate. By simple observation, negative coefficient will make the parabola graph should be concave down with maximum vertex value.

I got confused by the minus sign. My steps:
1. Multiplying the binomial, y=−(x²−6x−7)
2. Find x values at y=0, so 0=−(x²−6x−7)
3. Multiply both sides by -1 to remove the minus sign, so 0=(x²−6x−7)
4. factor them out, (x+1)(x−7) , so x=−1 and x=7
5. Transform the standard form into vertex form, y=(x²−6x+9)−9−7, then y=(x−3)²−16 ?? this step got me confused. with x=3 the vertex value will be y=−16 or in coordinate (3,−16) which is impossible!! because negative coefficient means the vertex should be positive not negative.

Was step number 3 a mistake? If i redo from step 3:
3. Put the minus sign in effect, 0=−x²+6x+7
4. Factor them out, (x−1)(x+7), so x=1 and x=−7 ?? i got different solutions??
5. Transform to vertex form, y=(−x²+6x−9)+9−7, then y=−(x−3)²+2 ?? I arrive at different vertex form. This case, i got vertex coordinate: (3,2) ?? This also impossible!! because both solutions were -7 and 1. it was impossible to have the vertex at x=3 !! vertex x-coordinate should be between both solutions.

Please end this minus sign trolling... i have become nuts...

 

[Ishuda]

I have quadratic functions: y=−(x+1)(x−7).
The goal is to find its solutions and vertex coordinate. By simple observation, negative coefficient will make the parabola graph should be concave down with maximum vertex value.

I got confused by the minus sign. My steps:
1. Multiplying the binomial, y=−(x²−6x−7)
2. Find x values at y=0, so 0=−(x²−6x−7)
3. Multiply both sides by -1 to remove the minus sign, so 0=(x²−6x−7)
4. factor them out, (x+1)(x−7) , so x=−1 and x=7
5. Transform the standard form into vertex form, y=(x²−6x+9)−9−7, then y=(x−3)²−16 ?? this step got me confused. with x=3 the vertex value will be y=−16 or in coordinate (3,−16) which is impossible!! because negative coefficient means the vertex should be positive not negative.

Was step number 3 a mistake? If i redo from step 3:
3. Put the minus sign in effect, 0=−x²+6x+7
4. Factor them out, (x−1)(x+7), so x=1 and x=−7 ?? i got different solutions??
5. Transform to vertex form, y=(−x²+6x−9)+9−7, then y=−(x−3)²+2 ?? I arrive at different vertex form. This case, i got vertex coordinate: (3,2) ?? This also impossible!! because both solutions were -7 and 1. it was impossible to have the vertex at x=3 !! vertex x-coordinate should be between both solutions.

Please end this minus sign trolling... i have become nuts...

No, the first step number 3 was not a mistake. Your first steps 1 to 4 are good and your second step 4 is incorrect, the factored equation would be (x+1)(7-x) or -(x+1)(x-7).

The mistake lies in the first step 5. Compare your step 1 and step 5. In one case you have a negative coefficient for x and in the other it is positive. Looking at the original equation we have
y=-[x² − 6x − 7] = ...

EDIT: Also the part in red above is incorrect if you mean the y value at the vertex should be positive. The only thing the negative coefficient for x² in a quadratic means is that the vertex will be a maximum. However that maximum can be a either a positive or negative number.

 

[lookagain]

I have quadratic functions: y=−(x+1)(x−7).The goal is to find its solutions and vertex coordinate. By simple observation, negative coefficient will make the parabola graph should be concave down with maximum vertex value.I got confused by the minus sign. My steps:1. Multiplying the binomial, y=−(x²−6x−7)2. Find x values at y=0, so 0=−(x²−6x−7)3. Multiply both sides by -1 to remove the minus sign, so 0=(x²−6x−7)4. factor them out, (x+1)(x−7) , so x=−1 and x=7

When the graph of the parabola crosses the x-axis, the x-value of the vertex
is just the average of the x-values of the x-intercepts.

Here, the x-value of the vertex is (-1 + 7)/2 = 3.

The y-value of the vertex is -(3 + 1)(3 - 7) = -(4)(-4) = 16

The vertex is (3, 16), not (3, -16).

 

[Suwandy]

Thanks to Ishuda and lookagain for the replies!

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No, the first step number 3 was not a mistake. Your first steps 1 to 4 are good and your second step 4 is incorrect, the factored equation would be (x+1)(7-x) or -(x+1)(x-7).

On my second step 4, the minus sign on x² confused me. Does that mean turning 0=−(x²−6x−7) into 0=−x²+6x+7 was a bad idea? Does having negative coefficient messed up the factoring process??
Because i thought factoring result of −x²+6x+7 was (x−1)(x+7), thus getting x=1 and x=−7 was already correct in process (not result). please correct me if i'm wrong.

The mistake lies in the first step 5. Compare your step 1 and step 5. In one case you have a negative coefficient for x and in the other it is positive. Looking at the original equation we have
y=-[x² − 6x − 7] = ...

On first step 5, i keep the function already multiplied by -1 that's why the original negative coefficient disappeared in first step 5. Again, does multiplying by -1 to the original function messed up the whole process???

EDIT: Also the part in red above is incorrect if you mean the y value at the vertex should be positive. The only thing the negative coefficient for x² in a quadratic means is that the vertex will be a maximum. However that maximum can be a either a positive or negative number.

Oh yeah! now that i give it a thought again, i think you are right on this! maximum can also be negative. Thank you for pointing this out for me.

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When the graph of the parabola crosses the x-axis, the x-value of the vertex
is just the average of the x-values of the x-intercepts.

Here, the x-value of the vertex is (-1 + 7)/2 = 3.

The y-value of the vertex is -(3 + 1)(3 - 7) = -(4)(-4) = 16

The vertex is (3, 16), not (3, -16).

I'm aware that the vertex x-value is only average of both x solutions. But i tend to try other approach to clarify whether my findings correct or incorrect. Because i believe different approach use same concept thus yielding same result (correct me if i'm wrong).
But when i found that i can't get vertex y-value = 16 with other approach, i was totally lost.

 

[Ishuda]

...
Because i thought factoring result of −x²+6x+7 was (x−1)(x+7), thus getting x=1 and x=−7 was already correct in process (not result). please correct me if i'm wrong.

−x²+6x+7 is not (x−1)(x+7), it is -(x−1)(x+7). For finding zeros, it doesn't make a difference since -1*0 is still zero.

On first step 5, i keep the function already multiplied by -1 that's why the original negative coefficient disappeared in first step 5. Again, does multiplying by -1 to the original function messed up the whole process??? ...

However, when dealing with the function itself, you need to keep track of that minus sign (or any other factors you might have factored out) so that, in this case,
-y = (x−3)²−16
or
y = -(x−3)²+16