Question

33. (a)
Draw the graph of p(x) = x2 + 3x - 4. Using the graph find the zeroes. Justify your answer.

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Answer 1

$Given \: p(x) = x^{2} + 3x - 4$

$Let \: y = p(x) = x^{2} + 3x - 4$

$\begin {tabular} {|c|c|c|c|c|c|} </p><p>\cline {1-6} x&-4&-2&0&1&2 \\</p><p>\cline {1-6} x^{2}&16&4&0&1&4 \\</p><p>\cline {1-6} 3x&-12&-6&0&3&6 \\</p><p>\cline {1-6} -4&+4&-4&-4&-4&-4 \\</p><p>\cline {1-6} y& 0&-6&-4&0&6 \\</p><p>\cline {1-6} (x,y)&(-4,0)&(-2,-6)&(0,-4)&(1,0)&(2,6) \\</p><p>\cline {1-6}</p><p>\end {tabular}$

Draw the points on a graph and joining them with a smooth curve , we get a parabola.

The parabola intersects x - axis at (-4,0 ) and (1,0) .

So, the zeroes are -4 and 1 .

Verification:

x² + 3x - 4 = 0

/* Splitting the middle term,we get */

=> x² + 4x - 1x - 4 = 0

=> x( x + 4 ) - 1( x + 4 ) = 0

=> (x + 4 )(x - 1) = 0

=> x + 4 = 0 Or x - 1 = 0

=> x = -4 Or x = 1

i.e., the zeroes of the given polynomial are -4 and -1

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