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    I need help with this question... Can someone please explain part a), i know that we draw each line alone but the second part of this function (x^2 +3) has no answer anyway (gives math error on calculator) !! So how did they obtain that curve for sec one I'm so confused .
    Also why did they substitute in the 1 I don't get that..
    Please help x
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    (Original post by pondsteps)
    I need help with this question... Can someone please explain part a), i know that we draw each line alone but the second part of this function (x^2 +3) has no answer anyway (gives math error on calculator) !! So how did they obtain that curve for sec one I'm so confused .
    Also why did they substitute in the 1 I don't get that..
    Please help x
    What do you mean it has no answer??? You simply sketch x^2+3 on the domain x\in [1,\infty)

    They substituted 1 to see what the minimum value of function is and begin sketching the part of the parabola from there.
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    (Original post by pondsteps)
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    I need help with this question... Can someone please explain part a), i know that we draw each line alone but the second part of this function (x^2 +3) has no answer anyway (gives math error on calculator) !! So how did they obtain that curve for sec one I'm so confused .
    Also why did they substitute in the 1 I don't get that..
    Please help x
    Hi,
    x^2 + 3 is just x^2 translated 3 up. You then only draw the graph for the given domain,
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    (Original post by RDKGames)
    What do you mean it has no answer??? You simply sketch x^2+3 on the domain x\in [1,\infty)

    They substituted 1 to see what the minimum value of function is and begin sketching the part of the parabola from there.
    Yeah but I usually do -b+-√(b^2 -4 x a x c) / 2a to find the values but It gave zeros this time so how am I supposed to sketch it ...

    Also is the one a random number?? Or does it have to be one that's substituted
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    (Original post by 111davey1)
    Hi,
    x^2 + 3 is just x^2 translated 3 up. You then only draw the graph for the given domain,
    Ohh right thank u x
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    (Original post by pondsteps)
    Yeah but I usually do -b+-√(b^2 -4 x a x c) / 2a to find the values but It gave zeros this time so how am I supposed to sketch it ...

    Also is the one a random number?? Or does it have to be one that's substituted
    You do realise that the application of the quadratic formula gives you the roots of the quadratic, ie the points on the x-axis where there parabola crosses it? That is not needed here. Clearly x^2+3 will never cross the x-axis as it is a translation vertically upwards from the basic parabola x^2 which barely touches the x-axis.

    You dont need to find the roots which are evidently not real. You simply need to know from what (x,y) point the curve begins, and you find that by plugging in x=1 as that is the minimum value your parabola can take within f(x) and then proceed to sketch the rest of it from that point for x\geq 1
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    (Original post by RDKGames)
    You do realise that the application of the quadratic formula gives you the roots of the quadratic, ie the points on the x-axis where there parabola crosses it? That is not needed here. Clearly x^2+3 will never cross the x-axis as it is a translation vertically upwards from the basic parabola x^2 which barely touches the x-axis.

    You dont need to find the roots which are evidently not real. You simply need to know from what (x,y) point the curve begins, and you find that by plugging in x=1 as that is the minimum value your parabola can take within f(x) and then proceed to sketch the rest of it from that point for x\geq 1
    Ohhh okay ! But I have one more question.. How could we find the min value of the curve when it has no real roots (nothing to do with this question) because the -b/2a will equal zero as the b is zero but the actual min value isn't zero as there is a y axis intersection at 3 .. So does this mean that the y axis interaction IS the min value (vertex) when there is no real roots??
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    (Original post by pondsteps)
    Ohhh okay ! But I have one more question.. How could we find the min value of the curve when it has no real roots (nothing to do with this question) because the -b/2a will equal zero as the b is zero but the actual min value isn't zero as there is a y axis intersection at 3 .. So does this mean that the y axis interaction IS the min value (vertex) when there is no real roots??
    Depends on what type of curve, in general you can find min/max points via differentiation of the function. For a quadratic you can either differentiate or complete the square.

    ax^2+bx+c=0 has completed square form of (x-\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2}=0 and since ax^2+bx+c is just a transformation from x^2 we can say that x^2 \mapsto ax^2+bx+c thus since the min point of x^2 is (0,0) then that gets mapped to (\frac{b}{2a},\frac{c}{a}-\frac{b^2}{4a^2}) via some transformations.

    For your question though it's different because it's not a full parabola so you need to have knowledge of how it would look like and where the minimum point would be then approach to finding it via substitution of an appropriate value.

    P.S. A function doesn't need to have real roots in order to have a minimum value... These are different things.
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    (Original post by RDKGames)
    Depends on what type of curve, in general you can find min/max points via differentiation of the function. For a quadratic you can either differentiate or complete the square.

    ax^2+bx+c=0 has completed square form of (x-\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2}=0 and since ax^2+bx+c is just a transformation from x^2 we can say that x^2 \mapsto ax^2+bx+c thus since the min point of x^2 is (0,0) then that gets mapped to (\frac{b}{2a},\frac{c}{a}-\frac{b^2}{4a^2}) via some transformations.

    For your question though it's different because it's not a full parabola so you need to have knowledge of how it would look like and where the minimum point would be then approach to finding it via substitution of an appropriate value.

    P.S. A function doesn't need to have real roots in order to have a minimum value... These are different things.
    So completeing the square gives us the min value of x????? I thought it gives the roots of the function!!!! 😭

    And also doesn't -b/2a give us the min point?? When do we use this

    Can u please explain in details .. Thanks
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    (Original post by pondsteps)
    So completeing the square gives us the min value of x????? I thought it gives the roots of the function!!!! 😭

    And also doesn't -b/2a give us the min point?? When do we use this

    Can u please explain in details .. Thanks
    Completing the square can be used for 2 purposes: finding the minimum/maximum point of the quadratic once it is expressed in the form (x-a)^2+b=0, or it can be used to find the roots by rearranging this form for x.

    What you've done was apply the quadratic formula which ONLY gives you roots which aren't always the minimum point.

    Also yes -\frac{b}{2a} gives the x-coordinate of the minimum point (I had a typo in the post above) and the y coordinate is given by \frac{c}{a}-\frac{b^2}{4a^2}
 
 
 
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